1 Introduction↩︎

To describe aggregation phenomena in bacterial populations, Keller and Segel ([1]) proposed the model \[\label{00} \left\{ \begin{array}{l} u_t = D \Delta u - \nabla\cdot (u\nabla v), \\[1mm] v_t = d \Delta v - v + u, \end{array} \right.\tag{1}\] with \(u=u(x,t)\) and \(v=v(x,t)\) denoting the population density and signal concentration, respectively, in which the bacteria are attracted by a chemical signal produced by themselves. Chemotaxis mechanisms of this form have been found to have wide applications in biology and ecology, and aslo in social sciences (cf., e.g., [2]).
This model (1 ) possesses two favorable mathematical properties: Firstly, it has a Lyapunov structure formally expressed in the identity \[\label{001} \frac{d}{dt} \bigg\{ \frac{d}{2} \int_\Omega|\nabla v|^2+\frac{1}{2}\int_\Omega v^2 -\int_\Omega uv +D\int_\Omega u \ln u\bigg\} =-\int_\Omega v_t^2 -\int_\Omega\Big| D\frac{\nabla u}{\sqrt{u}}-\sqrt{u}\nabla v\Big|^2,\tag{2}\] valid along suitably regular trajectories ([3]). Secondly, the attractant concentration \(v\) satisfies an inhomogeneous linear parabolic equation, accessible to classical analysis based on smoothing properties of corresponding heat semigroups. Suitable combination of these fundamental features has substantially influenced previous studies on (1 ), and has thereby, inter alia, facilitated the discovery of dichotomies between globally smooth behavior on the one hand ([3], [4]), and the occurrence of singularity formation on the other ([5], [6], [7], [8]); In addition to these and partially even farther-reaching results obtained for certain parabolic-elliptic simplifications ([9], [10], [11], [12], [13], [14], [15]), taxis-driven blow-up has also beed detected in closely related complex models (cf. [16], [17], [18], [19], [20], [21], [22] and [23] for a small selection).
A Keller-Segel-type model with anisotropic production of signals. In the study of clustering and pattern formation among autophoretic colloids, the authors in [24] found that the chemical is produced by the colloid asymmetrically due to the anisotropic properties of Janus particles, and they introduced an additional term \(\nabla\cdot (u\nabla v)\) to describe a certain anisotropic correction to the isotropic signal production term \(+u\). As a consequence, the signal evolution is accordingly characterized by an equation of the form \[\label{002} v_t = d \Delta v +\nabla\cdot (u\nabla v)- v + u.\tag{3}\] Although a number of variants of (1 ) that involve various alternative types of migration mechanisms such as nonlinear diffusion or modified chemotactic responses have been extensively studied in the literature ([25]), possible effects of such anisotropies in chemoattractant production have been much less explored so far. In fact, when embedded into the corresponding initial-boundary value problem \[\label{0} \left\{ \begin{array}{ll} u_t = D \Delta u - \nabla\cdot (u\nabla v), \qquad & x\in\Omega, \;t>0, \\[1mm] v_t = d \Delta v + \nabla\cdot (u\nabla v) - v + u, \qquad & x\in\Omega, \;t>0, \\[1mm] \frac{\partial u}{\partial\nu}=\frac{\partial v}{\partial\nu}=0, \qquad & x\in\partial\Omega, \;t>0, \\[1mm] u(x,0)=u_0(x), \quad v(x,0)=v_0(x), \qquad & x\in\Omega, \end{array} \right.\tag{4}\] to be subsequently considered in a smoothly bounded domain \(\Omega\subset\mathbb{R}^n\), this anisotropic correction term \(\nabla\cdot (u\nabla v)\) in the second equation brings about two analytical obstacles: It does not only destroy the Lypunov functional structure (2 ) for the original Keller-Segel model; beyond this, the principal part in the second subsystem of (4 ) thereby loses its linear structure and even contains a diffusion degeneracy that potentially might counteract higher-order regularity properties.
In line with this, the corresponding analytical literature so far seems limited to the recent study [26] in which (4 ) is examined for \(n\le 5\) and under the restriction that the difference \(|D-d|\) of the linear parts in both diffusion mechanisms is suitably small. Within this framework, by means of a non-symmetrically coupled gradient estimate technique along with a self-mapping argument and a refined Hölder regularity analysis a result on global existence of bounded classical solutions is derived in [26]. Although this markedly distinguishes (4 ) from the classical Keller-Segel model (1 ) with its well-known core property to generate blow-up when \(n\ge 2\), it leaves open the question how far the introduction of anisotropic corrections in signal production prevents chemotactic collapse also in the presence of strongly different diffusion rates; in fact, this question seems of particular relevance in cases in which, as typically seen in nature, individuals in the considered population move at velocities significantly smaller than signal molecules do.
Main results. The focus of the present work will accordingly be set on the development of a basic solution theory for (4 ) in settings of arbitrary \(D>0\) and \(d>0\), where mainly for technical purposes we shall concentrate on the case when \(n\le 5\). In order to circumvent obstacles linked to the diffusion degeneracy in the second equation from (4 ), our analysis in this regard will be based on a variational approach concentration on the evolution of spatial integrals that exclusively involve zero-order expressions. In order to nevertheless achieve suitably far-reaching information, the core part of our considerations will trace functionals of the form \[\label{003} \int_\Omega v^2 e^{(w+1)^\alpha} + b \int_\Omega(w+1) e^{(w+1)^\alpha}, \qquad w:=u+v, \qquad b>0,\tag{5}\] along trajectories (see Lemma 21, Lemma 25 and Lemma 26).
A priori estimates accordingly implied for solutions to certain regularized variants of (4 ) (see (23 )) will not only lead to a statement on global existence within a fairly natural notion of weak solvability, but furthermore provide time-independent bounds for \(u\) in an Orlicz class smaller than \(L^p(\Omega)\) for each finite \(p\); in particular, our following main result rules out any \(L^p\) blow-up phenomenon in (4 ) both in finite or in infinite time, contrary to the situation in the multi-dimensional version of (1 ) in which finite-time explosions actually occur in each of the spaces \(L^p(\Omega)\) with \(p>\frac{n}{2}\) when \(n\ge 2\) ([27], [6], [25]):

2 Preliminaries↩︎

2.1 Two families of interpolation inequalities↩︎

A key role in our analysis will be played by two functional inequalities which can be viewed as far relatives of Ehrling’s inequality, and which will be decisive in appropriately estimating zero-order expressions related to the source term \(+u\) appearing in the second equation in (4 ). In view of our ambition to subsequently consider solutions to the approximate variants of (4 ) introduced in (23 ) below, these inequalities will need to suitably cope with the appearance of a regularization parameter \(\varepsilon\) therein, and with consequences thereof on a reduced strength of the diffusion mechanism determining the evolution of the second solution component.
The first of these inequalities will be used in revealing quasi-energy properties enjoyed by certain combinations of functionals that exhibit essentially cubic growth with respect to both solution components (see Lemma 12):

Proof. An interpolation relying on the compactness of the embedding \(W^{1,2}(\Omega) \hookrightarrow L^2(\Omega)\) yields \(c_1=c_1(\eta)>0\) such that \[\label{1.2} \|\psi\|_{L^2(\Omega)}^2 \le \frac{4\eta}{9} \|\nabla\psi\|_{L^2(\Omega)}^2 + c_1 \|\psi\|_{L^\frac{2}{3}(\Omega)}^2 \qquad for all\psi\in C^1(\overline{\Omega}).\tag{6}\] Noting that for \[\begin{align} \rho_\varepsilon(\xi):=\frac{\sqrt{\xi+1}^3}{\sqrt{1+\varepsilon\xi}}, \qquad \xi\ge 0, \;\varepsilon\in (0,1), \end{align}\] we have \[\begin{align} \rho_\varepsilon'(\xi) &=& \frac{3}{2} \cdot \frac{\sqrt{\xi+1}}{\sqrt{1+\varepsilon\xi}} - \frac{\varepsilon}{2} \cdot \frac{\sqrt{\xi+1}^3}{\sqrt{1+\varepsilon\xi}^3} \\ &=& \frac{1}{2} \cdot \sqrt{\frac{\xi+1}{1+\varepsilon\xi}} \cdot \frac{3-\varepsilon+2\varepsilon\xi}{1+\varepsilon\xi} \qquad for all \xi\ge 0 and\varepsilon\in (0,1) \end{align}\] and hence \[\begin{align} 0\le \rho_\varepsilon'(\xi) \le \frac{1}{2} \cdot \sqrt{\frac{\xi+1}{1+\varepsilon\xi}} \cdot \frac{3+3\varepsilon\xi}{1+\varepsilon\xi} = \frac{3}{2} \cdot \sqrt{\frac{\xi+1}{1+\varepsilon\xi}} \qquad for all \xi\ge 0 and\varepsilon\in (0,1), \end{align}\] for fixed nonnegative \(\varphi\in C^1(\overline{\Omega})\) we obtain from (6 ) that \[\begin{align} \int_\Omega\frac{(\varphi+1)^3}{1+\varepsilon\varphi} &=& \big\| \rho_\varepsilon(\varphi)\big\|_{L^2(\Omega)}^2 \\ &\le& \frac{4\eta}{9} \big\| \nabla\rho_\varepsilon(\varphi)\big\|_{L^2(\Omega)}^2 + c_1 \big\| \rho_\varepsilon(\varphi)\big\|_{L^\frac{2}{3}(\Omega)}^2 \\ &=& \frac{4\eta}{9} \cdot \int_\Omega\rho_\varepsilon'^2(\varphi) |\nabla\varphi|^2 + c_1 \cdot \bigg\{ \int_\Omega\rho_\varepsilon^\frac{2}{3}(\varphi)\bigg\}^3 \\ &\le& \eta \int_\Omega\frac{\varphi+1}{1+\varepsilon\varphi} |\nabla\varphi|^2 + c_1 \cdot \bigg\{ \int_\Omega\rho_\varepsilon^\frac{2}{3}(\varphi)\bigg\}^3 \qquad for all\varepsilon\in (0,1). \end{align}\] Since \[\begin{align} c_1 \cdot \bigg\{ \int_\Omega\rho_\varepsilon^\frac{2}{3}(\varphi)\bigg\}^3 = c_1 \cdot \bigg\{ \int_\Omega\frac{\varphi+1}{(1+\varepsilon\varphi)^\frac{1}{3}} \bigg\}^3 \le c_1 \cdot \bigg\{ \int_\Omega(\varphi+1) \bigg\}^3 \qquad for all\varepsilon\in (0,1) \end{align}\] by nonnegativity of \(\varepsilon\varphi\) for any such \(\varepsilon\), this yields (?? ) with \(\Lambda_1(\eta):=c_1\). \(\Box\) Establishing a second and more subtle relation will require the following statement on zero-order interpolation as a preliminary.

Proof. We let \(N=N(\alpha)\ge 1\) be such that \[\begin{align} \alpha N^\alpha\ge 1-\alpha, \end{align}\] and note that then for \[\label{2.22} \rho(\xi):=(\xi+N)^\frac{\alpha-1}{2} e^{\frac{1}{2}(\xi+1)^\alpha}, \qquad \xi\ge 0,\tag{7}\] we have \[\begin{align} \label{2.23} \rho'(\xi) &=& \frac{\alpha}{2} (\xi+N)^\frac{\alpha-1}{2} (\xi+1)^{\alpha-1} e^{\frac{1}{2}(\xi+1)^\alpha} - \frac{1-\alpha}{2} (\xi+N)^\frac{\alpha-3}{2} e^{\frac{1}{2}(\xi+1)^\alpha} \nonumber\\ &=& \frac{1}{2} (\xi+N)^\frac{\alpha-3}{2} e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \big\{ \alpha(\xi+N)(\xi+1)^{\alpha-1} - (1-\alpha) \big\} \nonumber\\ &=& \frac{1}{2} (\xi+N)^\frac{\alpha-3}{2} e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \Big\{ \alpha(\xi+N)^\alpha\cdot \Big(\frac{\xi+N}{\xi+1}\Big)^{1-\alpha} - (1-\alpha)\Big\} \nonumber\\ &\ge& \frac{1}{2} (\xi+N)^\frac{\alpha-3}{2} e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \big\{ \alpha N^\alpha- (1-\alpha)\big\} \ge 0 \qquad for all\xi\ge 0, \end{align}\tag{8}\] because \(\alpha<1\). For \(0\le \varphi\in C^0(\overline{\Omega})\) fulfilling (?? ), and for arbitrary \(a>0\), splitting \[\label{2.3} \int_\Omega\frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} = \int_{\{\varphi<a\}} \frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} + \int_{\{\varphi\ge a\}} \frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} \qquad \varepsilon\in (0,1),\tag{9}\] we can therefore estimate \[\begin{align} \label{2.4} \int_{\{\varphi<a\}} \frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} &\le& \int_{\{\varphi<a\}} (\varphi+N)^\frac{\alpha+1}{2} e^{\frac{1}{2}(\varphi+1)^\alpha} \nonumber\\ &=& \int_{\{\varphi<a\}} (\varphi+N) \rho(\varphi) \nonumber\\ &\le& \rho(a) \int_{\{\varphi<a\}} (\varphi+N) \nonumber\\ &\le& (\mu+N|\Omega|) \rho(a) \qquad for all\varepsilon\in (0,1) \end{align}\tag{10}\] according to (?? ). Apart from that, simply using that \[\begin{align} \frac{(1+\varepsilon\varphi)^\frac{1}{2}}{(\varphi+1)^\frac{\alpha+1}{2}} \le \frac{(1+\varepsilon\varphi)^\frac{1}{2}}{(\varphi+1)^\frac{1}{2}} \le 1 \end{align}\] for \(\varepsilon\in (0,1)\), we see on writing \(I_\varepsilon(\varphi):=\int_\Omega\frac{(\varphi+1)^{\alpha+1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha}\), \(\varepsilon\in (0,1)\), that \[\begin{align} \int_{\{\varphi\ge a\}} \frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} &=& \int_{\{\varphi\ge a\}} \Big\{ \frac{(\varphi+1)^{\alpha+1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} \Big\} \cdot \frac{(1+\varepsilon\varphi)^\frac{1}{2}}{(\varphi+1)^\frac{\alpha+1}{2}} \cdot e^{-\frac{1}{2}(\varphi+1)^\alpha} \\ &\le& \int_{\{\varphi\ge a\}} \Big\{ \frac{(\varphi+1)^{\alpha+1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} \Big\} \cdot e^{-\frac{1}{2}(\varphi+1)^\alpha} \\[1mm] &\le& e^{-\frac{1}{2}(a+1)^\alpha} I_\varepsilon(\varphi) \qquad for all\varepsilon\in (0,1), \end{align}\] whence by (9 ) and (10 ), \[\label{2.5} \int_\Omega\frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} \le c_1 \rho(a) + e^{-\frac{1}{2}(a+1)^\alpha} I_\varepsilon(\varphi) \qquad for all \varepsilon\in (0,1) anda>0\tag{11}\] with \(c_1\equiv c_1(\alpha,\mu):=\mu+N|\Omega|\).
We now fix \(\eta>0\) and let \[\label{2.6} a_\varepsilon\equiv a_\varepsilon(\eta,\varphi):= \bigg\{ 2 \ln_+ \sqrt{\frac{4 I_\varepsilon(\varphi)}{\eta}} \bigg\}^\frac{1}{\alpha}, \qquad \varepsilon\in (0,1),\tag{12}\] where \(\ln_+ \xi:=\max\{0 \, , \, \ln \xi\}\) for \(\xi>0\). Then in the case when \(\varepsilon\in (0,1)\) is such that \[\label{2.66} a_\varepsilon\le a_0\equiv a_0(\eta,\alpha,\mu):= \Big(\frac{4c_1\sqrt{e}}{\eta}\Big)^\frac{2}{1-\alpha},\tag{13}\] we evidently have \(\ln \frac{4 I_\varepsilon(\varphi)}{\eta} < a_0^\alpha\), that is, \[\begin{align} I_\varepsilon(\varphi) \le \frac{\eta}{4} e^{a_0^\alpha}, \end{align}\] so that (11 ) together with (8 ) guarantees that \[\label{2.7} \int_\Omega\frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} \le c_2\equiv c_2(\eta,\alpha,\mu):= c_1 \rho(a_0) + e^{-\frac{1}{2}(a_0+1)^\alpha} \cdot \frac{\eta}{4} e^{a_0^\alpha}.\tag{14}\] If, conversely, \[\label{2.8} a_\varepsilon>a_0,\tag{15}\] then by (12 ), \[\label{2.81} e^{-\frac{1}{2}(a_\varepsilon+1)^\alpha} I_\varepsilon(\varphi) \le e^{-\frac{1}{2}a_\varepsilon^\alpha} I_\varepsilon(\varphi) = e^{-\ln \sqrt{\frac{4 I_\varepsilon(\varphi)}{\eta}}} I_\varepsilon(\varphi) = \frac{\sqrt{\eta}}{2} \cdot \sqrt{I_\varepsilon(\varphi)},\tag{16}\] while according to our definition of \(a_0\) in (13 ), and again since \(\alpha< 1\), we may estimate \((a_\varepsilon+1)^\alpha\le a_\varepsilon^\alpha+ 1\) to see that \[\begin{align} \label{2.82} c_1 \rho(a_\varepsilon) &=& c_1 (a_\varepsilon+N)^\frac{\alpha-1}{2} e^{\frac{1}{2}(a_\varepsilon+1)^\alpha} \nonumber\\ &\le& c_1 \sqrt{e} a_0^\frac{\alpha-1}{2} e^{\frac{1}{2} a_\varepsilon^\alpha} \nonumber\\ &=& c_1\sqrt{e} a_0^\frac{\alpha-1}{2} \cdot \sqrt{\frac{4I_\varepsilon(\varphi)}{\eta}} \nonumber\\ &=& \frac{\sqrt{\eta}}{2} \cdot \sqrt{I_\varepsilon(\varphi)}. \end{align}\tag{17}\] In view of (16 ) and (17 ), an application of (11 ) shows that whenever (15 ) holds, we have \[\begin{align} \bigg\{ \int_\Omega\frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2} (\varphi+1)^\alpha} \bigg\}^2 \le \Big\{ \frac{\sqrt{\eta}}{2} \cdot \sqrt{I_\varepsilon(\varphi)} + \frac{\sqrt{\eta}}{2} \cdot \sqrt{I_\varepsilon(\varphi)} \Big\}^2 = \eta I_\varepsilon(\varphi), \end{align}\] which in conjunction with (14 ) shows that (?? ) is valid for any choice of \(\varepsilon\in (0,1)\) if we let \(\Lambda_2(\eta,\alpha,\mu):=c_2^2\). \(\Box\) A second preparation consists in the following elementary observation.

Proof. We compute \[\label{31.3} \rho_\varepsilon'(\xi) = \frac{\alpha}{2} \cdot \frac{(\xi+1)^\frac{3\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} + \frac{\alpha+1}{2} \cdot \frac{(\xi+1)^\frac{\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} - \frac{\varepsilon}{2} \cdot \frac{(\xi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\xi)^\frac{3}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha}, \qquad \xi\ge 0,\tag{18}\] and thus obtain on dropping the nonnegative first summand here that \[\begin{align} \rho_\varepsilon'(\xi) &\ge& \frac{(\xi+1)^\frac{\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \Big\{ \frac{\alpha+1}{2} - \frac{\varepsilon}{2} \cdot \frac{\xi+1}{1+\varepsilon\xi} \Big\} \\ &\ge& \frac{(\xi+1)^\frac{\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \Big\{ \frac{\alpha+1}{2} - \frac{\varepsilon}{2} \cdot \frac{\xi+\frac{1}{\varepsilon}}{1+\varepsilon\xi} \Big\} \\ &=& \frac{(\xi+1)^\frac{\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \frac{\alpha}{2} \qquad for all\xi\ge 0, \end{align}\] which particularly yields the left inequality in (?? ). To verify the right one, we first observe that if \(\xi\ge 0\) is such that \((\xi+1)^\alpha\ge \frac{\alpha+1}{\alpha}\), then trivially estimating the rightmost summand in (18 ) we see that \[\begin{align} \rho_\varepsilon'(\xi) &\le& \frac{(\xi+1)^\frac{3\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \Big\{ \frac{\alpha}{2} + \frac{\alpha+1}{2} \cdot \frac{1}{(\xi+1)^\alpha} \Big\} \\ &\le& \frac{(\xi+1)^\frac{3\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \Big\{ \frac{\alpha}{2} + \frac{\alpha+1}{2} \cdot \frac{\alpha}{\alpha+1} \Big\} \\ &=& \frac{(\xi+1)^\frac{3\alpha-1}{2}}{(1+\varepsilon\xi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \cdot \alpha. \end{align}\] Since, on the other hand, for any \(\xi\ge 0\) fulfilling \((\xi+1)^\alpha< \frac{\alpha+1}{\alpha}\) we have \[\begin{align} \rho_\varepsilon'(\xi) &\le& \frac{\alpha}{2} \cdot (\xi+1)^\frac{3\alpha}{2} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} + \frac{\alpha+1}{2} \cdot (\xi+1)^\frac{\alpha}{2} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \\ &\le& \frac{\alpha}{2} \cdot \Big(\frac{\alpha+1}{\alpha}\Big)^\frac{3}{2} \cdot e^\frac{\alpha+1}{2\alpha} + \frac{\alpha+1}{2} \cdot\Big(\frac{\alpha+1}{\alpha}\Big)^\frac{1}{2} \cdot e^\frac{\alpha+1}{2\alpha} \\ &=& \Big(\frac{\alpha+1}{\alpha}\Big)^\frac{1}{2} \cdot \Big\{ \frac{\alpha}{2} \cdot \frac{\alpha+1}{\alpha} + \frac{\alpha+1}{2}\Big\} \cdot e^\frac{\alpha+1}{2\alpha}, \end{align}\] rearranging shows that (?? ) holds in both these cases. \(\Box\) Indeed, we can thereby derive a family of relatives of (?? ) which contain some superalgebraically growing quantities, and which will thereby form a crucial ingredient to our analysis related to the bounds claimed in (?? ), as to be detailed in Lemma 26.

Proof. We let \(q:=\max\{\frac{2n}{n+2},1\} \in [1,2)\), and may then rely on the continuity of the embedding \(W^{1,q}(\Omega) \hookrightarrow L^2(\Omega)\) to find \(c_1>0\) such that \[\label{32.2} \|\psi\|_{L^2(\Omega)}^2 \le c_1 \|\nabla\psi\|_{L^q(\Omega)}^2 + c_1 \|\psi\|_{L^1(\Omega)}^2 \qquad for all\psi\in C^1(\overline{\Omega}).\tag{19}\] Fixing \(\alpha\in (0,\min\{1,\frac{2}{n} \}]\), \(\mu>0\) and \(0\le\varphi\in C^1(\overline{\Omega})\) such that \(\int_\Omega\varphi\le \mu\), we thus obtain that if we let \((\rho_\varepsilon)_{\varepsilon\in (0,1)}\) be as in Lemma 4, then \[\begin{align} \label{32.3} \int_\Omega\frac{(\varphi+1)^{\alpha+1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} = \|\rho_\varepsilon(\varphi)\|_{L^2(\Omega)}^2 \le c_1\|\nabla\rho_\varepsilon(\varphi)\|_{L^q(\Omega)}^2 + c_1 \|\rho_\varepsilon(\varphi)\|_{L^1(\Omega)}^2 \end{align}\tag{20}\] for all \(\varepsilon\in (0,1)\). Here, abbreviating \(c_2\equiv c_2(\alpha):=\frac{(\alpha+1)^\frac{3}{2}}{\sqrt{\alpha}} \cdot e^\frac{\alpha+1}{2\alpha}\) we see that thanks to (?? ) and the Hölder inequality, \[\begin{align} \label{32.4} c_1 \|\nabla\rho_\varepsilon(\varphi)\|_{L^q(\Omega)}^2 &=& c_1 \|\rho_\varepsilon'(\varphi)\nabla\varphi\|_{L^q(\Omega)}^2 \nonumber\\ &\le& c_1 \cdot \bigg\| \Big\{ \alpha\cdot \frac{(\varphi+1)^\frac{3\alpha-1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\varphi+1)^\alpha} + c_2 \big\} \cdot |\nabla\varphi| \bigg\|_{L^q(\Omega)}^2 \nonumber\\ &\le& 2c_1 \alpha^2 \cdot \bigg\| \frac{(\varphi+1)^\frac{3\alpha-1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\varphi+1)^\alpha} |\nabla\varphi| \bigg\|_{L^q(\Omega)}^2 + 2c_1 c_2^2 \|\nabla\varphi\|_{L^q(\Omega)}^2 \nonumber\\ &=& 2c_1 \alpha^2 \cdot \bigg\{ \int_\Omega\Big\{ \frac{(\varphi+1)^{2\alpha-1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} |\nabla\varphi|^2 \Big\}^\frac{q}{2} \cdot (\varphi+1)^\frac{q\alpha}{2} \bigg\}^\frac{2}{q} + 2c_1 c_2^2 \cdot \bigg\{ \int_\Omega|\nabla\varphi|^q \bigg\}^\frac{2}{q} \nonumber\\ &\le& 2c_1 \alpha^2 \cdot \bigg\{ \int_\Omega\frac{(\varphi+1)^{2\alpha-1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} |\nabla\varphi|^2 \bigg\} \cdot \bigg\{ \int_\Omega(\varphi+1)^\frac{q\alpha}{2-q}\bigg\}^\frac{2-q}{q} \nonumber\\ & & + 2c_1 c_2^2 |\Omega|^\frac{2-q}{q} \int_\Omega|\nabla\varphi|^2 \qquad for all\varepsilon\in (0,1), \end{align}\tag{21}\] where we note that when \(n\ge 2\), we have \(q=\frac{2n}{n+2}\) and thus \(\frac{q\alpha}{2-q}=\frac{n\alpha}{2}\le 1\) due to our restriction that \(\alpha\le \frac{2}{n}\), while if \(n=1\), then \(q=1\) and hence \(\frac{q\alpha}{2-q}=\alpha\le 1\). Therefore, regardless of the size of \(n\) we may draw on (?? ) to estimate \[\begin{align} \int_\Omega(\varphi+1)^\frac{q\alpha}{2-q} \le \int_\Omega(\varphi+1) \le \mu+|\Omega|, \end{align}\] so that (21 ) ensures that \[\begin{align} \label{32.5} c_1 \|\nabla\rho_\varepsilon(\varphi)\|_{L^q(\Omega)}^2 &\le& 2c_1 \alpha^2 \cdot (\mu+|\Omega|)^\frac{2-q}{q} \int_\Omega\frac{(\varphi+1)^{2\alpha-1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} |\nabla\varphi|^2 \nonumber\\ & & + 2c_1 c_2^2 |\Omega|^\frac{2-q}{q} \int_\Omega|\nabla\varphi|^2 \qquad for all\varepsilon\in (0,1). \end{align}\tag{22}\] Since, apart from that, an application of Lemma 3 shows that if we let \(c_3\equiv c_3(\alpha,\mu):=\Lambda_2(\frac{1}{2c_1},\alpha,\mu)\) with \(\Lambda_2(\cdot,\cdot,\cdot)\) as found there, then \[\begin{align} c_1 \|\rho_\varepsilon(\varphi)\|_{L^1(\Omega)}^2 &=& c_1 \cdot \bigg\{ \int_\Omega\frac{(\varphi+1)^\frac{\alpha+1}{2}}{(1+\varepsilon\varphi)^\frac{1}{2}} \cdot e^{\frac{1}{2}(\varphi+1)^\alpha} \bigg\}^2 \\ &\le& \frac{1}{2} \int_\Omega\frac{(\varphi+1)^{\alpha+1}}{1+\varepsilon\varphi} \cdot e^{(\varphi+1)^\alpha} + c_3 \qquad for all\varepsilon\in (0,1), \end{align}\] a combination of (20 ) with (22 ) shows that (?? ) holds if we let \(\Lambda_3 \equiv \Lambda_3(\mu):=4c_1 \cdot (\mu+|\Omega|)^\frac{2-q}{q}\) and \(\Lambda_4\equiv \Lambda_4(\alpha,\mu):=\max\big\{ 4c_1 c_2^2 |\Omega|^\frac{2-q}{q} \, , \, 2c_3\big\}\). \(\Box\)

2.2 A quantitative outcome of a Moser-type iteration↩︎

The following is a consequence of a slightly more general statement on the outcome of a Moser-type iterative reasoning, as recorded in [28].

As substantiated in Lemma 12 below, our subsequent analysis will make use of Lemma 6 through the following consequence thereof.

Proof. Letting \(\widehat{z}:=z+1\), from (?? ) we obtain that \(\frac{\partial\widehat{z}}{\partial\nu}\le 0\) on \(\partial\Omega\times (0,T)\), and that since both \(f\) and \(z\) are nonnegative, \[\begin{align} \widehat{z}_t = z_t \le \nabla\cdot \big(a(x,t)\nabla z\big) + f(x,t) = \nabla\cdot \big(a(x,t)\nabla\widehat{z}\big) + \frac{1}{z+1} \cdot f(x,t)\widehat{z} \le \nabla\cdot \big(a(x,t)\nabla\widehat{z}\big) + f(x,t)\widehat{z} \end{align}\] in \(\Omega\times (0,T)\). Relying on (?? ) and (?? ), an application of Lemma 6 therefore shows that \[\begin{align} \|\widehat{z}(\cdot,t)\|_{L^\infty(\Omega)} \le \Lambda_4(q,L) \cdot \max \bigg\{ \|\widehat{z}(\cdot,0)\|_{L^\infty(\Omega)} \, , \, \sup_{s\in (0,T)} \|\widehat{z}(\cdot,s)\|_{L^1(\Omega)} \bigg\} \qquad for allt\in (0,T), \end{align}\] from which (?? ) already follows due to the fact that \(|z|=z\le\widehat{z}\) by nonnegativity of \(z\). \(\Box\)

3 An approximate variant of (4 ) and basic properties↩︎

As will turn out below, a regularization of (4 ) which does not only admit global classical solutions, but which simultaneously also is compatible with some favorable structural properties formally enjoyed by (4 ), to be discovered in Lemma 23 and Lemma 26, can be achieved by considering the family of problems given by \[\label{0eps} \left\{ \begin{array}{ll} u_{\varepsilon t} = D \Delta u_\varepsilon- \nabla\cdot\Big( \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \nabla v_\varepsilon\Big), \qquad & x\in\Omega, \;t>0, \\[1mm] v_{\varepsilon t} = d\Delta v_\varepsilon+ \nabla\cdot\Big( \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \nabla v_\varepsilon\Big) - v_\varepsilon+ \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}, \qquad & x\in\Omega, \;t>0, \\[1mm] \frac{\partial u_\varepsilon}{\partial\nu}=\frac{\partial v_\varepsilon}{\partial\nu}=0, \qquad & x\in\partial\Omega, \;t>0, \\[1mm] u_\varepsilon(x,0)=u_0(x), \quad v_\varepsilon(x,0)=v_0(x), \qquad & x\in\Omega, \end{array} \right.\tag{23}\] for \(\varepsilon\in (0,1)\). As a first step toward verifying this, let us record the following statement on local existence and extensibility therefor, and on two basic properties related to the evolution of corresponding mass functionals.

Proof. The statement concerning local existence and the extensibility criterion in (?? ) directly follows from the standard parabolic theory developed in [29], while the mass property (?? ) readily results from straightforward integration in (23 ) along with a simple ODE comparsion argument. \(\Box\) From now on, we shall fix \(n\le 5\) and a bounded domain \(\Omega\subset\mathbb{R}^n\) with smooth boundary, as well as numbers \(D>0\) and \(d>0\) and functions \(u_0\) and \(v_0\) fulfilling (?? ), and given \(\varepsilon\in (0,1)\) we then let \(T_{max,\varepsilon}\) and \((u_\varepsilon,v_\varepsilon)\) be as obtained in Lemma 8.
Our derivation of a first regularity property beyond those in (?? ) will make essential use of the circumstance that the regularization underlying (23 ) treats the crucial ingredients \(\pm \nabla\cdot (u\nabla v)\) to (4 ) in a synchronous manner, thus facilitating a favorable cancellation encountered when adding both parabolic equations in (23 ):

Proof. In view of Lemma 8, this can be seen by combining the two sub-problems of (23 ) in a straightforward manner. \(\Box\) The plain structure of (?? ) allows for simple testing procedures, a general template for which is recorded in the following.

Proof. Let \(\varepsilon\in (0,1)\). An integration by parts on the basis of (?? ) then shows that \[\begin{align} \label{33.2} \frac{d}{dt} \int_\Omega\rho(w_\varepsilon) &=& \int_\Omega\rho'(w_\varepsilon) \nabla\cdot \big\{ D\nabla w_\varepsilon+ (d-D)\nabla v_\varepsilon\big\} + \int_\Omega\rho'(w_\varepsilon) \cdot \Big\{ - v_\varepsilon+ \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \Big\} \nonumber\\ &=& - D \int_\Omega\rho''(w_\varepsilon) |\nabla w_\varepsilon|^2 - (d-D) \int_\Omega\rho''(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ & & - \int_\Omega v_\varepsilon\rho'(w_\varepsilon) + \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \cdot \rho'(w_\varepsilon) \qquad for allt\in (0,T_{max,\varepsilon}). \end{align}\tag{24}\] Here, relying on the nonnegativity of \(\rho''\) we may invoke Young’s inequality to see that for all \(t\in (0,T_{max,\varepsilon})\) we have \[\label{33.3} - (d-D) \int_\Omega\rho''(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon \le \frac{D}{2} \int_\Omega\rho''(w_\varepsilon) |\nabla w_\varepsilon|^2 + \frac{(d-D)^2}{2D} \int_\Omega\rho''(w_\varepsilon)|\nabla v_\varepsilon|^2,\tag{25}\] while using that \(u_\varepsilon\le w_\varepsilon\) and that \(\rho'\ge 0\) we can estimate \[\label{33.4} - \int_\Omega v_\varepsilon\rho'(w_\varepsilon) + \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \cdot \rho'(w_\varepsilon) \le \int_\Omega u_\varepsilon\rho'(w_\varepsilon) \le \int_\Omega(w_\varepsilon+1) \rho'(w_\varepsilon) \qquad for allt\in (0,T_{max,\varepsilon}).\tag{26}\] Inserting (25 ) and (26 ) into (24 ) yields (?? ). \(\Box\)

Proof. We differentiate to see that \[\label{4.23} \rho_\varepsilon'(\xi)=\frac{3(\xi+1)^2}{1+\varepsilon\xi} - \frac{\varepsilon(\xi+1)^3}{(1+\varepsilon\xi)^2} \qquad and \qquad \rho_\varepsilon''(\xi)=\frac{6(\xi+1)}{1+\varepsilon\xi} - \frac{6\varepsilon(\xi+1)^2}{(1+\varepsilon\xi)^2} + \frac{2\varepsilon^2(\xi+1)^3}{(1+\varepsilon\xi)^3}\tag{27}\] for all \(\xi\ge 0\), and that thus (?? ) follows upon observing that \[\begin{align} \frac{\varepsilon(\xi+1)^3}{(1+\varepsilon\xi)^2} = \frac{(\xi+1)^2}{1+\varepsilon\xi} \cdot \frac{\varepsilon\xi+\varepsilon}{1+\varepsilon\xi} \le \frac{(\xi+1)^2}{1+\varepsilon\xi} \qquad for all\xi\ge 0. \end{align}\] Since \[\begin{align} \frac{6\varepsilon(\xi+1)^2}{(1+\varepsilon\xi)^2} \le \frac{2\varepsilon^2(\xi+1)^3}{(1+\varepsilon\xi)^3} + \frac{9(\xi+1)}{2(1+\varepsilon\xi)} \qquad for all\xi\ge 0 \end{align}\] by Young’s inequality, and since \[\begin{align} \frac{\frac{2\varepsilon^2(\xi+1)^3}{(1+\varepsilon\xi)^3}}{\frac{6(\xi+1)}{1+\varepsilon\xi}} = \frac{\varepsilon^2(\xi+1)^2}{3(1+\varepsilon\xi)^2} = \frac{(\varepsilon+\varepsilon\xi)^2}{3(1+\varepsilon\xi)^2} \le \frac{1}{3} \qquad for all\xi\ge 0, \end{align}\] from (27 ) we moerover obtain (?? ). \(\Box\) In conjunction with an inequality describing the evolution of \(t\mapsto \int_\Omega(v_\varepsilon+1)^3\) for \(\varepsilon\in (0,1)\) (see (29 ), a particular version of (?? ) can be used to establish a first collection of estimates which can be viewed as approximate counterparts of corresponding properties addressed in [26] for the unperturbed problem (4 ):

Proof. For \(\varepsilon\in (0,1)\) we let \(\rho_\varepsilon\) be as in Lemma 11, and drawing on the left inequalities in (?? ) and (?? ) we may employ Lemma 10 to see that, again thanks to (?? ) and (?? ), \[\begin{align} \label{4.3}\frac{d}{dt} \int_\Omega\rho_\varepsilon(w_\varepsilon) &\le& - \frac{D}{2} \int_\Omega\rho_\varepsilon''(w_\varepsilon) |\nabla w_\varepsilon|^2 + \frac{(d-D)^2}{2D} \int_\Omega\rho_\varepsilon''(w_\varepsilon) |\nabla v_\varepsilon|^2 + \int_\Omega(w_\varepsilon+1)\rho_\varepsilon'(w_\varepsilon) \nonumber\\ &\le& - \frac{3D}{4} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla w_\varepsilon|^2 + \frac{4(d-D)^2}{D} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla v_\varepsilon|^2 + 3 \int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} \end{align}\tag{28}\] for all \(t\in (0,T_{max,\varepsilon})\). We next abbreviate \(d_0:=\min\{d,1\}\) and observe that since \(u_\varepsilon\le w_\varepsilon\) and thus \(\frac{1}{1+\varepsilon u_\varepsilon} \ge \frac{1}{1+\varepsilon w_\varepsilon}\), \[\begin{align} \Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \cdot (v_\varepsilon+1) &\ge& \Big(d+\frac{u_\varepsilon}{1+\varepsilon w_\varepsilon}\Big) \cdot (v_\varepsilon+1) \\ &=& \frac{(d+d\varepsilon w_\varepsilon+u_\varepsilon)(v_\varepsilon+1)}{1+\varepsilon w_\varepsilon} \\ &\ge& \frac{(d+u_\varepsilon)(v_\varepsilon+1)}{1+\varepsilon w_\varepsilon} \\ &=& \frac{dv_\varepsilon+d+u_\varepsilon v_\varepsilon+u_\varepsilon}{1+\varepsilon w_\varepsilon} \\ &\ge& \frac{dv_\varepsilon+d+u_\varepsilon}{1+\varepsilon w_\varepsilon} \\ &\ge& \frac{d_0 v_\varepsilon+ d_0 + d_0u_\varepsilon}{1+\varepsilon w_\varepsilon} \\ &=& d_0 \cdot \frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \quad in\Omega\times (0,T_{max,\varepsilon}). \end{align}\] Testing the seond equation in (23 ) by \((v_\varepsilon+1)^2\) we thus obtain that \[\begin{align} \label{4.4} & &\frac{1}{3} \frac{d}{dt} \int_\Omega(v_\varepsilon+1)^3 \nonumber\\ &=& \int_\Omega(v_\varepsilon+1)^2 \nabla\cdot \Big\{ \Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big)\nabla v_\varepsilon\Big\} - \int_\Omega(v_\varepsilon+1)^2 v_\varepsilon + \int_\Omega(v_\varepsilon+1)^2 \cdot \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \nonumber\\ &=& - 2 \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big)\cdot (v_\varepsilon+1) |\nabla v_\varepsilon|^2 - \int_\Omega(v_\varepsilon+1)^3 + \int_\Omega(v_\varepsilon+1)^2 \cdot \Big(1+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \nonumber\\ &\le& - 2 d_0 \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla v_\varepsilon|^2 - \int_\Omega(v_\varepsilon+1)^3 + \int_\Omega(v_\varepsilon+1)^2 \cdot \Big(1+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \end{align}\tag{29}\] for all \(t\in (0,T_{max,\varepsilon})\). Here, using that \(\frac{d}{d\xi} \frac{\xi}{1+\varepsilon\xi}\ge 0\) for all \(\xi\ge 0\) we can estimate \[\begin{align} 1+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \le 1 + \frac{w_\varepsilon}{1+\varepsilon w_\varepsilon} = \frac{1+(\varepsilon+1)w_\varepsilon}{1+\varepsilon w_\varepsilon} \le 2\cdot \frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \qquad in\Omega\times (0,T_{max,\varepsilon}), \end{align}\] and employ Young’s inequality to see that, accordingly, \[\begin{align} \int_\Omega(v_\varepsilon+1)^2 \cdot\Big(1+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) &\le& \frac{2}{3} \int_\Omega(v_\varepsilon+1)^3 + \frac{1}{3} \int_\Omega\Big(1+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big)^3 \\ &\le& \frac{2}{3} \int_\Omega(v_\varepsilon+1)^3 + \frac{8}{3} \int_\Omega\Big(\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon}\Big)^3 \\ &\le& \frac{2}{3} \int_\Omega(v_\varepsilon+1)^3 + \frac{8}{3} \int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} \qquad for allt\in (0,T_{max,\varepsilon}). \end{align}\] Writing \(b:=\frac{4(d-D)^2+D}{6d_0 D}\) and recalling (?? ), from (28 ) and (29 ) we hence infer that \[\begin{align} \label{4.5} & &\frac{d}{dt} \bigg\{ \int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} + b\int_\Omega(v_\varepsilon+1)^3 \bigg\} + \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla v_\varepsilon|^2 + \frac{3D}{4} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla w_\varepsilon|^2 \nonumber\\ & &+ \int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} + b\int_\Omega(v_\varepsilon+1)^3 \nonumber\\ &\le& (4+8b) \int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} \qquad for allt\in (0,T_{max,\varepsilon}). \end{align}\tag{30}\] Now Lemma 2 says that if we let \(c_1:=(4+8b)\Lambda_1\Big(\frac{3D}{8(4+8b)}\Big)\) with \(\Lambda_1(\cdot)\) as found there, then \[\begin{align} (4+8b)\int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} \le \frac{3D}{8} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla w_\varepsilon|^2 + c_1\cdot \bigg\{ \int_\Omega(w_\varepsilon+1)\bigg\}^3 \qquad for allt\in (0,T_{max,\varepsilon}), \end{align}\] so that since \[\begin{align} \int_\Omega(w_\varepsilon+1) = \int_\Omega u_\varepsilon+ \int_\Omega v_\varepsilon+ |\Omega| \le \int_\Omega u_0 + \max \bigg\{ \int_\Omega u_0 \, , \, \int_\Omega v_0\bigg\} + |\Omega| \le c_2\equiv c_2(K):=2K|\Omega| + |\Omega| \end{align}\] by (?? ), (?? ) and (?? ), we conclude that \[\begin{align} y_\varepsilon(t):=\int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} + b\int_\Omega(v_\varepsilon+1)^3, \qquad t\in [0,T_{max,\varepsilon}), \;\varepsilon\in (0,1), \end{align}\] as well as \[\begin{align} h_\varepsilon(t):= \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla v_\varepsilon|^2 + \frac{3D}{8} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} |\nabla w_\varepsilon|^2 \qquad t\in (0,T_{max,\varepsilon}), \;\varepsilon\in (0,1), \end{align}\] satisfy \[\label{4.99} y_\varepsilon'(t) + y_\varepsilon(t) + h_\varepsilon(t) \le c_1 c_2^3 \qquad for allt\in (0,T_{max,\varepsilon}).\tag{31}\] Again relying on (?? ), by means of an ODE comparison argument and an integration we obtain from this that \[\begin{align} y_\varepsilon(t) &\le& \max \bigg\{ c_1 c_2^3 \, , \, \int_\Omega\frac{(u_0+v_0+1)^3}{1+\varepsilon(u_0+v_0)} + b\int_\Omega(v_0+1)^3 \bigg\} \\ &\le& c_4\equiv c_4(K) := \max \Big\{ c_1 c_2^3 \, , \, (2K+1)^3 |\Omega| + b\cdot (K+1)^3 |\Omega| \Big\} \qquad for allt\in [0,T_{max,\varepsilon}), \end{align}\] and that thus, by an integration in (31 ), \[\begin{align} \int_t^{t+\tau_\varepsilon} h_\varepsilon(s) ds \le y_\varepsilon(t) + c_1 c_2^3 \tau_\varepsilon \le c_4 + c_1 c_2^3 \qquad for all t\in (0,T_{max,\varepsilon}-\tau_\varepsilon), \end{align}\] because \(\tau_\varepsilon\le 1\) for all \(\varepsilon\in (0,1)\). According to our definitions of \((y_\varepsilon)_{\varepsilon\in (0,1)}\) and \((h_\varepsilon)_{\varepsilon\in (0,1)}\), the claim thus readily follows upon observing that in line with (?? ) and the inequality \(u_\varepsilon\le w_\varepsilon\) we have \[\begin{align} \int_\Omega\Big(\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big)^3 \le \int_\Omega\frac{(u_\varepsilon+1)^3}{1+\varepsilon u_\varepsilon} = \int_\Omega\rho_\varepsilon(u_\varepsilon) \le \int_\Omega\rho_\varepsilon(w_\varepsilon) = \int_\Omega\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} \qquad for allt\in (0,T_{max,\varepsilon}), \end{align}\] and that \(\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \ge 1\) as well as \(\frac{(w_\varepsilon+1)^3}{1+\varepsilon w_\varepsilon} \ge w_\varepsilon^2\) due to the fact that \(\frac{\xi+1}{1+\varepsilon\xi}\ge 1\) for all \(\xi\ge 0\). \(\Box\) The actually most important implication of this section can now be achieved by drawing on the time-independent bound in \(L^3(\Omega)\) stated in (?? ) for the source term \(\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\) in the second equation in (23 ). Indeed, Corollary 7 says that in low-dimensional scenarios in which \(\frac{n}{2}<3\), this is sufficient to ensure \(L^\infty\) bounds for the \(v_\varepsilon\):

Proof. If (?? ) and (?? ) hold, then according to the nonnegativity of the \(v_\varepsilon\), (23 ) implies that \[\begin{align} v_{\varepsilon t} \le \nabla\cdot \big(a_\varepsilon(x,t)\nabla v_\varepsilon\big) + f_\varepsilon(x,t) \quad in\Omega\times (0,T_{max,\varepsilon}) \qquad for all\varepsilon\in (0,1), \end{align}\] where \(a_\varepsilon:=d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\) and \(f_\varepsilon:=\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\) satisfy \(a_\varepsilon\ge d\), \(f_\varepsilon\ge 0\) and, by (?? ), \[\begin{align} \int_\Omega f_\varepsilon^3 \le \Gamma_1 \qquad for all t\in (0,T_{max,\varepsilon}) and\varepsilon\in (0,1), \end{align}\] with \(\Gamma_1:=\Gamma_1(K)\) and \(\Gamma_1(\cdot)\) taken from Lemma 12. Using that \(3>\frac{n}{2}\), we may thus draw on Corollary 7 to see with with \(\Lambda_4(\cdot,\cdot)\) as introduced there, thanks to (?? ) and (?? ) we have \[\begin{align} \|v_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)} &\le& \Lambda_4\Big(3,\max\Big\{\frac{1}{d},\Gamma_1^\frac{1}{3} \Big\} \Big) \cdot \max \bigg\{ \|v_0+1\|_{L^\infty(\Omega)} \, , \, \sup_{s\in (0,T)} \|v(\cdot,s)+1\|_{L^1(\Omega)} \bigg\} \\ &\le& \Lambda_4\Big(3,\max\Big\{\frac{1}{d},\Gamma_1^\frac{1}{3} \Big\} \Big) \cdot (K+1)\max\{1, \, |\Omega|\} \end{align}\] for all \(t\in (0,T_{max,\varepsilon})\) and \(\varepsilon\in (0,1)\). \(\Box\)

4 Global solvability of (23 )↩︎

Unlike in the original problem (4 ), \(L^\infty\) bounds for the second components of the solutions to the non-degenerate regularized variants (23 ) already entail higher regularity features. The key toward this can be verified by reduction to standard literature on parabolic regularity theory:

Proof. The second equation in (23 ) can be recast according to \[\begin{align} v_{\varepsilon t}= \nabla\cdot \big( a_\varepsilon(x,t)\nabla v_\varepsilon\big) + f_\varepsilon(x,t), \qquad x\in\Omega, \;t\in (0,T_{max,\varepsilon}), \end{align}\] with \(a_\varepsilon:=d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\) and \(f_\varepsilon:=-v_\varepsilon+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\) satisfying \[\begin{align} d \le a_\varepsilon\le d+\frac{1}{\varepsilon} \quad and \quad |f_\varepsilon| \le M + \frac{1}{\varepsilon} \qquad in\Omega\times (0,T_{max,\varepsilon}) \end{align}\] by Lemma 13, where \(M=M(K)\) with \(K:=\|u_0\|_{L^\infty(\Omega)} + \|v_0\|_{L^\infty(\Omega)}\). Again explicitly relying on Lemma 13, we therefore obtain the claim as an immediate consequence of known results on Hölder regularity of bounded solutions to scalar parabolic problems ([30]). \(\Box\) When combined with (?? ), the information on time-independent Hölder regularity of the \(v_\varepsilon\) contained in (?? ) can be seen to imply an \(L^\infty\) bound for the first component in (23 ):

Proof. Lemma 12 guarantees the existence of \(c_1=c_1(u_0, v_0)>0\) such that \[\label{60.2} \int_\Omega w_\varepsilon^2(\cdot, t) \le c_1 \qquad for all t\in (0,T_{max,\varepsilon}) and\varepsilon\in (0,1),\tag{32}\] while (?? ) provides \(c_2=c_2(\varepsilon, u_0, v_0)\) such that \[\begin{align} \|v_\varepsilon(\cdot, t)\|_{C^\theta(\overline{\Omega})}\le c_2 \qquad for all t\in (0,T_{max,\varepsilon}) and\varepsilon\in (0,1). \end{align}\] Together with [26], the latter entails the existence of \(c_3=c_3(D, \theta)>0\) satisfying \[\label{60.3} \Bigg\| \int_0^t e^{(t-s)(D\Delta-1)} \Delta v_\varepsilon(\cdot,s) ds \Bigg\|_{L^\infty(\Omega)} \le c_3\cdot\sup_{s\in (0,t)} \|v_\varepsilon(\cdot,s)\|_{C^\theta(\overline{\Omega})} \le c_3\cdot c_2\tag{33}\] for all \(t\in (0,T_{max,\varepsilon})\) and \(\varepsilon\in (0,1)\). Since the first equation in (?? ) can be rewritten in the form \[\begin{align} w_{\varepsilon t} = D\Delta w_\varepsilon-w_\varepsilon+ (d-D) \Delta v_\varepsilon- v_\varepsilon+ \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} +w_\varepsilon, \qquad & x\in\Omega, \;t\in (0,T_{max,\varepsilon}), \end{align}\] relying on order preservation of \((e^{\tau\Delta})_{\tau\ge 0}\) and noting \(\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \le u_\varepsilon\le w_\varepsilon\) we obtain from an associatd Duhamel representation that \[\begin{align} w_\varepsilon(\cdot,t) &=& e^{t(D\Delta-1)} w_0 + (d-D) \int_0^t e^{(t-s)(D\Delta-1)} \Delta v_\varepsilon(\cdot,s) ds \\ & & - \int_0^t e^{(t-s)(D\Delta-1)} v_\varepsilon(\cdot,s) ds + \int_0^t e^{(t-s)(D\Delta-1)} \Big\{\frac{u_\varepsilon(\cdot, s)}{1+\varepsilon u_\varepsilon(\cdot, s)} +w_\varepsilon(\cdot,s)\Big\} ds \\ &\le& \|w_0\|_{L^\infty(\Omega)} + (d-D) \int_0^t e^{(t-s)(D\Delta-1)} \Delta v_\varepsilon(\cdot,s) ds + 2\int_0^t e^{(t-s)(D\Delta-1)} w_\varepsilon(\cdot,s) ds \quad in\Omega \end{align}\] for all \(t\in (0,T_{max,\varepsilon})\) and \(\varepsilon\in (0,1)\), so that from (33 ) we infer that for all \(t\in (0,T_{max,\varepsilon})\) and \(\varepsilon\in (0,1)\), \[\begin{align} \label{60.4} \|w_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)} &\le &\|w_0\|_{L^\infty(\Omega)} + (d-D) \cdot c_2 c_3 \nonumber\\ & & +2 \int_0^t \Big(1+D^{-\frac{n}{6}}(t-s)^{-\frac{n}{6}}\Big) e^{-(t-s)}\| w_\varepsilon(\cdot,s)\|_{L^3(\Omega)} ds. \end{align}\tag{34}\] For fixed \(T\in (0, T_{max,\varepsilon})\) writing \(A_\varepsilon(T):=\max_{t\in [0, T]} \|w_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)}\) and \(c_4=c_4(\varepsilon, u_0, v_0):=\|u_0+v_0\|_{L^\infty(\Omega)} + (d-D) \cdot c_2 c_3\), from (34 ) and a simple interpolation we obtain that due to (32 ), \[\begin{align} \|w_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)} &\le& c_4 + 2\int_0^t \Big(1+D^{-\frac{n}{6}}(t-s)^{-\frac{n}{6}}\Big) e^{-(t-s)}\|w_\varepsilon(\cdot,s)\|_{L^2(\Omega)}^\frac{2}{3} \cdot \|w_\varepsilon(\cdot,s)\|_{L^\infty(\Omega)}^\frac{1}{3} ds\\ &\le& c_4 +2c_1^\frac{1}{3}c_5\cdot A_\varepsilon^\frac{1}{3}(T) \qquad for allt\in (0, T), \end{align}\] where \(c_5:=\int_0^\infty \big(1+D^{-\frac{n}{6}}\sigma^{-\frac{n}{6}}\big) e^{-\sigma} d\sigma\) is finite according to our assumption that \(n\le 5\). In conjunction with Young’s inequality, this entails that \[\begin{align} A_\varepsilon(T)\le c_4 +2c_1^\frac{1}{3}c_5\cdot A_\varepsilon^\frac{1}{3}(T) \le c_4 +\frac{2}{3} A_\varepsilon(T) +\frac{4}{3} c_1^\frac{1}{2}c_5^\frac{3}{2} \end{align}\] and that, consequently, \[\begin{align} A_\varepsilon(T)\le C(\varepsilon, u_0, v_0):=3 c_4 +4 c_1^\frac{1}{2}c_5^\frac{3}{2} \qquad for allt\in (0, T), \end{align}\] which implies (?? ) due to the evident fact that \(\|w_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)} \ge \|u_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)}\) for all \(t\in (0,T_{max,\varepsilon})\). \(\Box\) Having the above information at hand, we can rearrange the approach developed for (4 ) in [26] to establish bounds for gradients of solutions to (23 ) in \(L^p\) spaces with arbitrarily large finite \(p\).

Proof. A proof is sketched in an appendix below. \(\Box\) In consequence, each of our approximate solutions actually is global in time:

Proof. In view of (?? ), this directly results from (?? ) and an application of Lemma 16 to any \(p\ge 4\) fulfilling \(p>n\). \(\Box\)

5 Constructing a global weak solution of (4 )↩︎

The mere construction of a global weak solution to (4 ) can, in its essence, already be based solely on Lemma 12 and the following fairly straightforward consequence thereof on time regularity.

Proof. For definiteness in notation, we fix \(c_1>0\) such that for each \(\psi\in C^1(\overline{\Omega})\) fulfilling \(\|\psi\|_{W^{1,6}(\Omega)} \le 1\) we have \(\|\nabla\psi\|_{L^6(\Omega)} + \|\nabla\psi\|_{L^2(\Omega)} + \|\psi\|_{L^\frac{3}{2}(\Omega)} + \|\psi\|_{L^1(\Omega)} \le c_1\). For any such \(\psi\), recalling that \(u_\varepsilon=w_\varepsilon-v_\varepsilon\) for \(\varepsilon\in (0,1)\), integrating by parts in (23 ), we then obtain that \[\begin{align} \bigg| \int_\Omega u_{\varepsilon t} \psi \bigg| &=& \bigg| - D \int_\Omega\nabla w_\varepsilon\cdot\nabla\psi + \int_\Omega\Big\{ D + \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \Big\} \nabla v_\varepsilon\cdot\nabla\psi \bigg| \\ &\le& D \|\nabla w_\varepsilon\|_{L^2(\Omega)} \|\nabla\psi\|_{L^2(\Omega)} + \Big\|D + \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\|_{L^3(\Omega)} \|\nabla v_\varepsilon\|_{L^2(\Omega)} \|\nabla\psi\|_{L^6(\Omega)} \\ &\le& c_1 D \|\nabla w_\varepsilon\|_{L^2(\Omega)} + c_1 \Big\|D + \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\|_{L^3(\Omega)} \|\nabla v_\varepsilon\|_{L^2(\Omega)} \qquad for all t>0 and\varepsilon\in (0,1) \end{align}\] and thus \[\begin{align} \label{81.3} & &\int_0^T \|u_{\varepsilon t}(\cdot,t)\|_{(W^{1,6}(\Omega))^\star}^2 dt \nonumber\\ &\le& \int_0^T \Big\{ c_1 D \|\nabla w_\varepsilon(\cdot,t)\|_{L^2(\Omega)} + c_1 \Big\| D +\frac{u_\varepsilon(\cdot,t)}{1+\varepsilon u_\varepsilon(\cdot,t)}\Big\|_{L^3(\Omega)} \|\nabla v_\varepsilon(\cdot,t)\|_{L^2(\Omega)} \Big\}^2 dt \nonumber\\ &\le& 2c_1^2 D^2 \int_0^T \int_\Omega|\nabla w_\varepsilon|^2 + 2c_1^2 \cdot \bigg\{ D|\Omega|^\frac{1}{3} + \sup_{t>0} \Big\|\frac{u_\varepsilon(\cdot,t)}{1+\varepsilon u_\varepsilon(\cdot,t)}\Big\|_{L^3(\Omega)} \bigg\}^2 \cdot \int_0^T \int_\Omega|\nabla v_\varepsilon|^2 \end{align}\tag{35}\] for all \(T>0\) and \(\varepsilon\in (0,1)\). Similarly, (23 ) implies that for all \(t>0\) and \(\varepsilon\in (0,1)\), \[\begin{align} \bigg| \int_\Omega v_{\varepsilon t} \psi \bigg| &=& \bigg| - \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \nabla v_\varepsilon\cdot\nabla\psi - \int_\Omega v_\varepsilon\psi + \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \psi \bigg| \\ &\le& \Big\| d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\|_{L^3(\Omega)} \|\nabla v_\varepsilon\|_{L^2(\Omega)} \|\nabla\psi\|_{L^6(\Omega)} \\ & & + \|v_\varepsilon\|_{L^\infty(\Omega)} \|\psi\|_{L^1(\Omega)} + \Big\|\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\|_{L^3(\Omega)} \|\psi\|_{L^\frac{3}{2}(\Omega)} \\ &\le& c_1 \cdot \Big\{ d |\Omega|^\frac{1}{3} + \Big\|\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\|_{L^3(\Omega)} \Big\} \cdot \|\nabla v_\varepsilon\|_{L^2(\Omega)} + c_1 \|v_\varepsilon\|_{L^\infty(\Omega)} + c_1 \Big\|\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\|_{L^3(\Omega)} \end{align}\] and hence \[\begin{align} \label{81.4} & &\int_0^T \|v_{\varepsilon t}(\cdot,t)\|_{(W^{1,6}(\Omega))^\star}^2 dt \nonumber\\ &\le& 3c_1^2 \cdot \bigg\{ d|\Omega|^\frac{1}{3} + \sup_{t>0} \Big\|\frac{u_\varepsilon(\cdot,t)}{1+\varepsilon u_\varepsilon(\cdot,t)}\Big\|_{L^3(\Omega)} \bigg\}^2 \cdot \int_0^T \int_\Omega|\nabla v_\varepsilon|^2 \nonumber\\ & & + 3c_1^2 \cdot \bigg\{ \sup_{t>0} \|v_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)} \bigg\}^2 \cdot T + 3c_1^2 \cdot \bigg\{ \sup_{t>0} \Big\|\frac{u_\varepsilon(\cdot,t)}{1+\varepsilon u_\varepsilon(\cdot,t)}\Big\|_{L^3(\Omega)} \bigg\}^2 \cdot T \end{align}\tag{36}\] for all \(T>0\) and \(\varepsilon\in (0,1)\). In view of Lemma 12 and Lemma 13, from (35 ) and (36 ) we obtain both (?? ) and (?? ) with some suitably large \(C(T,u_0,v_0)>0\). \(\Box\) Indeed, a combination of Lemma 12 with Lemma 18 yields the following.

Proof. For \(T>0\), from Lemma 12, (?? ) and (?? ) we know that \[\begin{align} (u_\varepsilon)_{\varepsilon\in (0,1)} \quad and \quad (v_\varepsilon)_{\varepsilon\in (0,1)} \quad are bounded inL^2((0,T);W^{1,2}(\Omega)), \end{align}\] while Lemma 18 asserts that \[\begin{align} (u_{\varepsilon t})_{\varepsilon\in (0,1)} \quad and \quad (v_{\varepsilon t})_{\varepsilon\in (0,1)} \quad are bounded inL^2\big((0,T);(W^{1,6}(\Omega))^\star\big). \end{align}\] Two applications of an Aubin-Lions lemma thus yield \((\varepsilon_j)_{j\in\mathbb{N}} \subset (0,1)\) as well as nonnegative elements \(u\) and \(v\) of \(L^2_{loc}([0,\infty);W^{1,2}(\Omega))\) such that \(\varepsilon_j\searrow 0\) as \(j\to\infty\), and that (?? )-(?? ) hold as \(\varepsilon=\varepsilon_j\searrow 0\). For the derivation of (?? ) and (?? ), we fix \(\varphi\in C_0^\infty(\overline{\Omega}\times [0,\infty))\) and then see on integrating by parts in (23 ) that \[\label{82.6} - \int_0^\infty \int_\Omega u_\varepsilon\varphi_t - \int_\Omega u_0 \varphi(\cdot,0) = -D \int_0^\infty \int_\Omega\nabla u_\varepsilon\cdot\nabla\varphi + \int_0^\infty \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \nabla v_\varepsilon\cdot\nabla\varphi\tag{37}\] as well as \[\begin{align} \label{82.7} - \int_0^\infty \int_\Omega v_\varepsilon\varphi_t - \int_\Omega v_0\varphi(\cdot,0) &=& - d \int_0^\infty \int_\Omega\nabla v_\varepsilon\cdot\nabla\varphi - \int_0^\infty \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \nabla v_\varepsilon\cdot\nabla\varphi\nonumber\\ & & - \int_0^\infty \int_\Omega v_\varepsilon\varphi + \int_0^\infty \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \varphi \end{align}\tag{38}\] for all \(\varepsilon\in (0,1)\). Now from (?? ) and the dominated convergence theorem it readily follows that \(\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \to u\) in \(L^2_{loc}(\overline{\Omega}\times [0,\infty))\) as \(\varepsilon=\varepsilon_j\searrow 0\), which combind with (?? ) shows that not only \[\begin{align} \int_0^\infty \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \varphi\to \int_0^\infty \int_\Omega u\varphi \qquad as \varepsilon=\varepsilon_j\searrow 0, \end{align}\] but also \[\begin{align} \int_0^\infty \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \nabla v_\varepsilon\cdot\nabla\varphi \to \int_0^\infty \int_\Omega u\nabla v\cdot\nabla\varphi \qquad as \varepsilon=\varepsilon_j\searrow 0. \end{align}\] Since clearly \[\begin{align} \int_0^\infty \int_\Omega u_\varepsilon\varphi_t \to \int_0^\infty \int_\Omega u\varphi_t, \quad \int_0^\infty \int_\Omega v_\varepsilon\varphi_t \to \int_0^\infty \int_\Omega v\varphi_t \quad and \quad \int_0^\infty \int_\Omega v_\varepsilon\varphi\to \int_0^\infty \int_\Omega v\varphi \end{align}\] as \(\varepsilon=\varepsilon_j\searrow 0\) by (?? ) and (?? ), and since moreover \[\begin{align} \int_0^\infty \int_\Omega\nabla u_\varepsilon\cdot\nabla\varphi\to \int_0^\infty \int_\Omega\nabla u\cdot\nabla\varphi \quad and \quad \int_0^\infty \int_\Omega\nabla v_\varepsilon\cdot\nabla\varphi\to \int_0^\infty \int_\Omega\nabla v\cdot\nabla\varphi \qquad as \varepsilon=\varepsilon_j\searrow 0 \end{align}\] due to (?? ) and (?? ), from (37 ) and (38 ) we infer that indeed both (?? ) and (?? ) hold. \(\Box\)

6 Controlling \(u_\varepsilon\) in exponential Orlicz classes. Proof of Theorem 1↩︎

6.1 An approximate variant of \(\int_\Omega(w+1)e^{(w+1)^\alpha}\) and its evolution↩︎

Next approaching the core of our analysis, in this part we will address an approximate counterpart of the Orlicz class estimate in (?? ). A first step toward this will rely on the outcome of Lemma 10 when applied to the functions introduced and characterized as follows.

Proof. Using that \(\frac{d}{d\xi} \frac{\xi+1}{1+\varepsilon\xi}=\frac{1-\varepsilon}{(1+\varepsilon\xi)^2}\) for all \(\xi\ge 0\), we calculate \[\label{333.5} \rho_\varepsilon'(\xi) = \alpha\cdot \frac{(\xi+1)^\alpha}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} + (1-\varepsilon) \cdot \frac{1}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha}\tag{39}\] and \[\begin{align} \label{333.6} \rho_\varepsilon''(\xi) &=& \alpha^2 \cdot \frac{(\xi+1)^{2\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} + \alpha^2 \cdot \frac{(\xi+1)^{\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} - \alpha\varepsilon\cdot \frac{(\xi+1)^\alpha}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha} \nonumber\\ & & + \alpha(1-\varepsilon) \cdot \frac{(\xi+1)^{\alpha-1}}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha} - 2\varepsilon(1-\varepsilon)\cdot \frac{1}{(1+\varepsilon\xi)^3} \cdot e^{(\xi+1)^\alpha} \end{align}\tag{40}\] for \(\xi\ge 0\), and observe that if \(\xi\ge 0\) is such that \[\label{333.7} (\xi+1)^\alpha\ge \frac{1}{\alpha},\tag{41}\] then \[\begin{align} \frac{(1-\varepsilon) \cdot \frac{1}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha}}{\alpha\cdot \frac{(\xi+1)^\alpha}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha}} = \frac{1-\varepsilon}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha(1+\varepsilon\xi)} \le \frac{1}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha} \le 1 \end{align}\] and thus \[\begin{align} (\xi+1) \rho_\varepsilon'(\xi) \le 2\alpha\cdot \frac{(\xi+1)^{\alpha+1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha}. \end{align}\] If \(\xi\ge 0\) is such that (41 ) does not hold, however, then \[\begin{align} (\xi+1) \cdot (1-\varepsilon) \cdot \frac{1}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha} \le (\xi+1) e^{(\xi+1)^\alpha} \le \Big(\frac{1}{\alpha}\Big)^\frac{1}{\alpha} \cdot e^\frac{1}{\alpha}, \end{align}\] so that (?? ) follows due to our assumption that \(\varepsilon<1\).
Likewise, for \(\xi\ge 0\) fulfilling (41 ) we can estimate \[\begin{align} \frac{\rho_\varepsilon(\xi)}{\alpha\cdot\frac{(\xi+1)^{\alpha+1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha}} = \frac{1}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha} \le 1, \end{align}\] while for \(\xi\ge 0\) satisfying \((\xi+1)^\alpha< \frac{1}{\alpha}\) we have \[\begin{align} \rho_\varepsilon(\xi) \le (\xi+1) e^{(\xi+1)^\alpha} \le \Big(\frac{1}{\alpha}\Big)^\frac{1}{\alpha} \cdot e^\frac{1}{\alpha}, \end{align}\] meaning that also (?? ) holds.
In quite a similar fashion, for \(\xi\ge 0\) we see that if (41 ) is valid, then \[\begin{align} \frac{\alpha(1-\varepsilon)\cdot\frac{(\xi+1)^{\alpha-1}}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha}}{\alpha^2 \cdot \frac{(\xi+1)^{2\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha}} = \frac{1-\varepsilon}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha(1+\varepsilon\xi)} \le \frac{1}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha} \le 1, \end{align}\] whereas otherwise, \[\begin{align} \alpha(1-\varepsilon) \cdot \frac{(\xi+1)^{\alpha-1}}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha} \le \alpha(\xi+1)^\alpha e^{(\xi+1)^\alpha} \le \alpha\cdot \frac{1}{\alpha} \cdot e^\frac{1}{\alpha} = e^\frac{1}{\alpha}. \end{align}\] Since moreover \((\xi+1)^{\alpha-1} \le (\xi+1)^{2\alpha-1}\) for all \(\xi\ge 0\) and thus \[\begin{align} \alpha^2 \cdot \frac{(\xi+1)^{\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} \le \alpha^2 \cdot \frac{(\xi+1)^{2\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} \qquad for all\xi\ge 0, \end{align}\] again relying on the fact that \(\varepsilon\in (0,1)\) we therefore obtain the right inequality in (?? ) from (40 ).
The claimed lower bound for \(\rho_\varepsilon''\), finally, can be verified by making use of our restriction in (?? ), which namely asserts that \[\begin{align} \frac{\alpha\varepsilon\cdot \frac{(\xi+1)^\alpha}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha}}{\alpha^2 \cdot \frac{(\xi+1)^{2\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha}} &=& \frac{\varepsilon}{\alpha} \cdot \frac{(\xi+1)^{1-\alpha}}{1+\varepsilon\xi} \\ &=& \frac{\varepsilon^\alpha}{\alpha} \cdot \frac{(\varepsilon\xi+\varepsilon)^{1-\alpha}}{(1+\varepsilon\xi)^{1-\alpha}} \cdot \frac{1}{(1+\varepsilon\xi)^\alpha} \\ &\le& \frac{\varepsilon^\alpha}{\alpha} \le \frac{1}{2} \qquad for all\xi\ge 0, \end{align}\] and that, similarly, \[\begin{align} \frac{2\varepsilon(1-\varepsilon)\cdot\frac{1}{(1+\varepsilon\xi)^3} \cdot e^{(\xi+1)^\alpha}}{\alpha(1-\varepsilon)\cdot \frac{(\xi+1)^{\alpha-1}}{(1+\varepsilon\xi)^2} \cdot e^{(\xi+1)^\alpha}} = 2\cdot\frac{\varepsilon}{\alpha} \cdot \frac{(\xi+1)^{1-\alpha}}{1+\varepsilon\xi} \le 1 \qquad for all\xi\ge 0. \end{align}\] Consequently, (40 ) implies that indeed also the left inequality in (?? ) holds. \(\Box\) Collecting the above list of inequalities shows that the general evolution property from Lemma 10 can be turned into the following starting point of our analysis toward (?? ).

Proof. We let \(\rho_\varepsilon\) be as in Lemma 20, and note that then \[\begin{align} \int_\Omega(w_\varepsilon+1) \rho_\varepsilon'(w_\varepsilon) \le 2\alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + \Big(\frac{e}{\alpha}\Big)^\frac{1}{\alpha} \cdot |\Omega| \qquad for allt>0 \end{align}\] by (?? ), and that \[\begin{align} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} = \int_\Omega\rho_\varepsilon(w_\varepsilon) \le \alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + \Big(\frac{e}{\alpha}\Big)^\frac{1}{\alpha} \cdot |\Omega| \qquad for allt>0 \end{align}\] thanks to (?? ). Therefore, (?? ) is a consequence of Lemma 10 when combined with (?? ). \(\Box\)

6.2 A functional involving a multiplicative coupling of \(v_\varepsilon\) and \(w_\varepsilon\)↩︎

Now the key step will aim at an appropriate control of the first integral on the right-hand side of (?? ), viewed here as an expression in the flavor of a Dirichlet integral over \(v_\varepsilon\) that involves a weight function depending on \(w_\varepsilon\) in a rapidly growing manner. Our approach toward a compensation of this will be based on an analysis of functionals which for \(\varepsilon\in (0,1)\) couple \(v_\varepsilon\) to \(w_\varepsilon\) in a certain multiplicative manner, allowing for some superalgebraic dependencies on \(w_\varepsilon\). An initial observation in this direction will be formulated in Lemma 23, making use of the simple two-sided estimate on the effective diffusion rate in the second equation in (23 ).

Proof. Let \(\varepsilon\in (0,1)\). Then again since \(\frac{d}{d\xi} \frac{\xi}{1+\varepsilon\xi} \ge 0\) for all \(\xi\ge 0\), the fact that \(u_\varepsilon\le w_\varepsilon\) implies that \[\begin{align} d + \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} &\le& d + \frac{w_\varepsilon}{1+\varepsilon w_\varepsilon} = \frac{d+d\varepsilon w_\varepsilon+w_\varepsilon}{1+\varepsilon w_\varepsilon} \le \frac{d+1+dw_\varepsilon+w_\varepsilon}{1+\varepsilon w_\varepsilon} \\ &=& (d+1) \cdot \frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \qquad in\Omega\times (0,\infty), \end{align}\] from which the right inequality in (?? ) follows.
We next rely on Lemma 13, which namely asserts that once more writing \(d_0:=\min\{d,1\}\) we have \(w_\varepsilon+1\le u_\varepsilon+M+1\) and hence, by nonnegativity of \(w_\varepsilon\) and \(u_\varepsilon\), \[\begin{align} \frac{d+\frac{u_\varepsilon}{1+\varepsilon w_\varepsilon}}{\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon}} &=& \frac{d+d\varepsilon w_\varepsilon+u_\varepsilon}{w_\varepsilon+1} \ge \frac{d+u_\varepsilon}{u_\varepsilon+M+1} \ge \frac{d_0+d_0u_\varepsilon}{u_\varepsilon+M+1} = d_0 - \frac{d_0 M}{u_\varepsilon+M+1} \\ &\ge& d_0 - \frac{d_0 M}{M+1} = \frac{d_0}{M+1} \qquad in\Omega\times (0,\infty). \end{align}\] This implies the left inequality in (?? ), because clearly \(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} \ge d + \frac{u_\varepsilon}{1+\varepsilon w_\varepsilon}\) in \(\Omega\times (0,\infty)\). \(\Box\) Relying on the latter in its technical part, the following lemma records the outcome of a procedure which in its essence can be viewed as consisting in a multiplication of the second equation in (23 ) by the product of \(v_\varepsilon\) with a function \(\chi(w_\varepsilon)\). It turns out that if here \(\chi\) satisfies a growth condition mild enough so as to be satisfied by functions of the form \(0\le\xi\mapsto e^{(\xi+1)^\alpha}\) for small \(\alpha>0\), then effects due to the cross-diffusive action expressed in (?? ) can be limited to the appearance of integrals exclusively involving \(w_\varepsilon\) and its gradient:

Proof. Given \(K>0\), we let \(M=M(K)>0\) be as in Lemma 13, and defining \[\label{11.2} \gamma\equiv \gamma(K):=\frac{\min\{d,1\}}{4\cdot (M+1)}\tag{42}\] we choose \(\alpha_0=\alpha_0(K)\in (0,1]\) in such a way that \[\label{11.22} 2|d-D| \cdot M \cdot \alpha_0 \le \gamma.\tag{43}\] Then assuming that (?? ) and (?? ) are valid, and that \(\chi\in C^2([0,\infty))\) is positive and satisfies (?? ), for fixed \(\varepsilon\in (0,1)\) we integrate by parts using (23 ) and (?? ) to compute \[\begin{align} \label{11.3} \frac{d}{dt} \int_\Omega v_\varepsilon^2 \chi(w_\varepsilon) &=& 2 \int_\Omega v_\varepsilon\chi(w_\varepsilon) \nabla\cdot \Big\{ \Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big)\nabla v_\varepsilon\Big\} + 2 \int_\Omega v_\varepsilon\chi(w_\varepsilon) \cdot \Big\{ - v_\varepsilon+ \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\} \nonumber\\ & & + \int_\Omega v_\varepsilon^2 \chi'(w_\varepsilon) \nabla\cdot \big\{ D\nabla w_\varepsilon+(d-D)\nabla v_\varepsilon\big\} + \int_\Omega v_\varepsilon^2 \chi'(w_\varepsilon) \cdot \Big\{ - v_\varepsilon+ \frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big\} \nonumber\\ &=& - 2 \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 - 2 \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) v_\varepsilon\chi'(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ & & - 2 \int_\Omega v_\varepsilon^2 \chi(w_\varepsilon) + 2 \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} v_\varepsilon\chi(w_\varepsilon) \nonumber\\ & & -2D \int_\Omega v_\varepsilon\chi'(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon - 2(d-D) \int_\Omega v_\varepsilon\chi'(w_\varepsilon) |\nabla v_\varepsilon|^2 \nonumber\\ & & - D \int_\Omega v_\varepsilon^2 \chi''(w_\varepsilon) |\nabla w_\varepsilon|^2 - (d-D) \int_\Omega v_\varepsilon^2 \chi''(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ & & - \int_\Omega v_\varepsilon^3 \chi'(w_\varepsilon)+ \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}v_\varepsilon^2 \chi'(w_\varepsilon) \qquad for allt>0. \end{align}\tag{44}\] Here in view of the positivity of \(\chi\), Young’s inequality together with Lemma 13 and the right inequality in (?? ) guarantees that \[\begin{align} & &- 2 \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) v_\varepsilon\chi'(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ &\le& \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) v_\varepsilon^2 \cdot \frac{\chi'^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \nonumber\\ &\le& \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + (d+1)M^2 \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \frac{\chi'^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \qquad for allt>0, \end{align}\] while thanks to the left inequality in (?? ), \[\begin{align} \int_\Omega\Big(d+\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big) \chi(w_\varepsilon) |\nabla w_\varepsilon|^2 \ge 4\gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 \qquad for allt>0. \end{align}\] As \(\chi'\ge 0\) and \[\begin{align} - D \int_\Omega v_\varepsilon^2 \chi''(w_\varepsilon) |\nabla w_\varepsilon|^2 \le D M^2 \int_{\{\chi''(w_\varepsilon)<0\}} |\chi''(w_\varepsilon)| \cdot |\nabla w_\varepsilon|^2 \qquad for allt>0 \end{align}\] by Lemma 13, from (44 ) we thus obtain that \[\begin{align} \label{11.4} & &\frac{d}{dt} \int_\Omega v_\varepsilon^2 \chi(w_\varepsilon) + 4\gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + 2 \int_\Omega v_\varepsilon^2 \chi(w_\varepsilon) \nonumber\\ &\le& (d+1)M^2 \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \frac{\chi'^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 + D M^2 \int_{\{\chi''(w_\varepsilon)<0\}} |\chi''(w_\varepsilon)| \cdot |\nabla w_\varepsilon|^2 \nonumber\\ & & -2D \int_\Omega v_\varepsilon\chi'(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon - 2(d-D) \int_\Omega v_\varepsilon\chi'(w_\varepsilon) |\nabla v_\varepsilon|^2 \nonumber\\ & & - (d-D) \int_\Omega v_\varepsilon^2 \chi''(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon + 2 \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} v_\varepsilon\chi(w_\varepsilon) + \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}v_\varepsilon^2 \chi'(w_\varepsilon) \end{align}\tag{45}\] for all \(t>0\), and here two further applications of Young’s inequality show that again due to Lemma 13, \[\begin{align} \label{11.5} & &-2D \int_\Omega v_\varepsilon\chi'(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ &\le& \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + \frac{D^2}{\gamma} \int_\Omega v_\varepsilon^2 \cdot \frac{1+\varepsilon w_\varepsilon}{w_\varepsilon+1} \cdot \frac{\chi'^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \nonumber\\ &\le& \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + \frac{D^2 M^2}{\gamma} \int_\Omega\frac{1+\varepsilon w_\varepsilon}{w_\varepsilon+1} \cdot \frac{\chi'^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \end{align}\tag{46}\] and \[\begin{align} \label{11.6} & &- (d-D) \int_\Omega v_\varepsilon^2 \chi''(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ &\le& \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + \frac{(d-D)^2}{4\gamma} \int_\Omega v_\varepsilon^4 \cdot \frac{1+\varepsilon w_\varepsilon}{w_\varepsilon+1} \cdot \frac{\chi''^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \nonumber\\ &\le& \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + \frac{(d-D)^2 M^4}{4\gamma} \int_\Omega\frac{1+\varepsilon w_\varepsilon}{w_\varepsilon+1} \cdot \frac{\chi''^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \end{align}\tag{47}\] for all \(t>0\). Noting that \(\frac{1+\varepsilon\xi}{\xi+1} \le 1\) for all \(\xi\ge 0\), we may estimate \[\label{11.66} \frac{1+\varepsilon w_\varepsilon}{w_\varepsilon+1} \le 1 \le \frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \qquad in\Omega\times (0,\infty)\tag{48}\] to see that (46 ) and (47 ) imply that for all \(t>0\), \[\begin{align} \label{11.7} & &-2D \int_\Omega v_\varepsilon\chi'(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon - (d-D) \int_\Omega v_\varepsilon^2 \chi''(w_\varepsilon) \nabla v_\varepsilon\cdot\nabla w_\varepsilon\nonumber\\ &\le& 2 \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 + \frac{D^2 M^2}{\gamma} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \frac{\chi'^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 \nonumber\\ & & + \frac{(d-D)^2 M^4}{4\gamma} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \frac{\chi''^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2. \end{align}\tag{49}\] Apart from that, we may control the fourth to last summand in (45 ) by combining (?? ) with (43 ), according to which, namely, it follows that again due to Lemma 13 and (48 ), \[\begin{align} \label{11.8} - 2(d-D) \int_\Omega v_\varepsilon\chi'(w_\varepsilon) |\nabla v_\varepsilon|^2 &\le& 2|d-D| \cdot M \int_\Omega\chi'(w_\varepsilon) |\nabla v_\varepsilon|^2 \nonumber\\ &\le& 2|d-D| \cdot M \cdot \alpha_0 \int_\Omega\chi(w_\varepsilon) |\nabla v_\varepsilon|^2 \nonumber\\ &\le& 2|d-D| \cdot M \cdot \alpha_0 \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 \nonumber\\ &\le& \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \chi(w_\varepsilon) |\nabla v_\varepsilon|^2 \qquad for allt>0. \end{align}\tag{50}\] Since, finally, \[\begin{align} 2 \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon} v_\varepsilon\chi(w_\varepsilon) + \int_\Omega\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}v_\varepsilon^2 \chi'(w_\varepsilon) &\le& (2M +\alpha_0 M^2) \int_\Omega\frac{w_\varepsilon}{1+\varepsilon w_\varepsilon} \cdot \chi(w_\varepsilon) \\ &\le& (2M +\alpha_0 M^2) \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \chi(w_\varepsilon) \qquad for allt>0 \end{align}\] by Lemma 13 and (?? ) as well as the upward monotonicity of \(0\le\xi\mapsto \frac{\xi}{1+\varepsilon\xi}\), from (45 ), (49 ) and (50 ) we readily infer that (?? ) holds if we let \(\Gamma(K):=\max \Big\{ (d+1)M^2 + \frac{D^2 M^2}{\gamma}, \, \frac{(d-D)^2 M^4}{4\gamma}, \, D M^2,\) \(2M +\alpha_0 M^2 \Big\}\). \(\Box\) Now the particular structure of the first integral on the right of (?? ) suggests to here choose the function \(\chi\) to be a member of the family characterized in the following lemma.

Proof. Differentiating in (?? ) yields (?? ) as well as the identity \[\label{35.7} \chi''(\xi) = \alpha^2 (\xi+1)^{2\alpha-2} e^{(\xi+1)^\alpha} - \alpha(1-\alpha) (\xi+1)^{\alpha-2} e^{(\xi+1)^\alpha} \qquad for all\xi\ge 0,\tag{51}\] where in the case when \(\xi\ge 0\) satisfies \((\xi+1)^\alpha\ge\frac{2}{\alpha}\), we see that \[\begin{align} \frac{\alpha(1-\alpha)(\xi+1)^{\alpha-2} e^{(\xi+1)^\alpha}}{\alpha^2 (\xi+1)^{2\alpha-2} e^{(\xi+1)^\alpha}} = \frac{1-\alpha}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha} \le \frac{1-\alpha}{2} \le \frac{1}{2} \end{align}\] and hence, in particular, \[\label{35.8} \frac{\alpha^2}{2} (\xi+1)^{2\alpha-2} e^{(\xi+1)^\alpha} \le \chi''(\xi) \le \alpha^2 (\xi+1)^{2\alpha-2} e^{(\xi+1)^\alpha} \qquad for all \xi\ge 0 fulfilling (\xi+1)^\alpha\ge \frac{2}{\alpha}.\tag{52}\] On the other hand, for any \(\xi\ge 0\) satisfying \((\xi+1)^\alpha<\frac{2}{\alpha}\) we have \[\label{35.9} \alpha(1-\alpha)(\xi+1)^{\alpha-2} e^{(\xi+1)^\alpha} \le \alpha(\xi+1)^\alpha e^{(\xi+1)^\alpha} \le \alpha\cdot\frac{2}{\alpha} \cdot e^\frac{2}{\alpha} = 2e^\frac{2}{\alpha}\tag{53}\] and, apart from that, \[\begin{align} \label{35.10} \sqrt{\frac{\xi+1}{1+\varepsilon\xi}} \cdot \frac{\alpha(1-\alpha)(\xi+1)^{\alpha-2} e^{(\xi+1)^\alpha}}{\sqrt{\chi(\xi)}} &=& \alpha(1-\alpha)\cdot \frac{(\xi+1)^{\alpha-\frac{3}{2}}}{\sqrt{1+\varepsilon\xi}} \cdot e^{\frac{1}{2}(\xi+1)^\alpha} \nonumber\\ &\le& \alpha\cdot (\xi+1)^\alpha\cdot e^{\frac{1}{2}(\xi+1)^\alpha} \nonumber\\ &\le& \alpha\cdot\frac{2}{\alpha} \cdot e^\frac{1}{\alpha} = 2 e^\frac{1}{\alpha} \qquad for all\varepsilon\in (0,1). \end{align}\tag{54}\] Now (53 ) together with (51 ) shows that \[\begin{align} \chi''(\xi) \ge \alpha^2(\xi+1)^{2\alpha-2} e^{(\xi+1)^\alpha} - 2e^\frac{2}{\alpha} \qquad for all \xi\ge 0 such that (\xi+1)^\alpha<\frac{2}{\alpha}, \end{align}\] which combined with (52 ) establishes (?? ).
Apart from that, (54 ) along with (52 ) and (51 ) implies that whenever \(\xi\ge 0\) is such that \(\chi''(\xi)\le 0\), \[\label{35.11} \frac{\xi+1}{1+\varepsilon\xi} \cdot \frac{\chi''^2(\xi)}{\chi(\xi)} \le 4 e^\frac{2}{\alpha} \qquad for all\varepsilon\in (0,1),\tag{55}\] while within \(\{\chi''>0\}\) it follows from (51 ) and the inequality \(\alpha\le 1\) that \[\begin{align} \label{35.12} \frac{\xi+1}{1+\varepsilon\xi} \cdot \frac{\chi''^2(\xi)}{\chi(\xi)} &\le& \frac{\xi+1}{1+\varepsilon\xi} \cdot \frac{\big\{ \alpha^2 (\xi+1)^{2\alpha-2} e^{(\xi+1)^\alpha} \big\}^2}{e^{(\xi+1)^\alpha}} \nonumber\\ &=& \alpha^4\cdot \frac{(\xi+1)^{4\alpha-3}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} \nonumber\\ &\le& \alpha^2 \cdot \frac{(\xi+1)^{2\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} \qquad for all\varepsilon\in (0,1). \end{align}\tag{56}\] As clearly \[\begin{align} \frac{\xi+1}{1+\varepsilon\xi} \cdot \frac{\chi'^2(\xi)}{\chi(\xi)} = \alpha^2 \cdot \frac{(\xi+1)^{2\alpha-1}}{1+\varepsilon\xi} \cdot e^{(\xi+1)^\alpha} \qquad for all \xi\ge 0 and\varepsilon\in (0,1) \end{align}\] by (?? ), from (55 ) and (56 ) we infer (?? ) for arbitrary \(\varepsilon\in (0,1)\).
Finally, given \(\xi\ge 0\) we obtain from (?? ) that if \((\xi+1)^\alpha\ge\frac{1}{\alpha}\), then \[\begin{align} \frac{\frac{\xi+1}{1+\varepsilon\xi} \cdot \chi(\xi)}{\alpha\cdot\frac{(\xi+1)^{\alpha+1}}{1+\varepsilon\xi}\cdot e^{(\xi+1)^\alpha}} = \frac{1}{\alpha} \cdot \frac{1}{(\xi+1)^\alpha} \le 1 \qquad for all\varepsilon\in (0,1), \end{align}\] while if \((\xi+1)^\alpha< \frac{1}{\alpha}\), then \[\begin{align} \frac{\xi+1}{1+\varepsilon\xi} \cdot\chi(\xi) \le (\xi+1) e^{(\xi+1)^\alpha} \le \Big(\frac{1}{\alpha}\Big)^\frac{1}{\alpha} \cdot e^\frac{1}{\alpha} \qquad for all\varepsilon\in (0,1). \end{align}\] This confirms (?? ), whereas (?? ) can directly be derived from (?? ) by trivially estimating \((\xi+1)^\alpha\le (\xi+1)^{\alpha+1}\) for \(\xi\ge 0\). \(\Box\) Indeed, when spelt out for functions of this form, Lemma 23 leads to the main result of this section:

Proof. We let \(\chi\) be as defined in Lemma 24 and note that then, by (?? ), \[\begin{align} 0 \le\chi'(\xi) \le \alpha(\xi+1)^{\alpha-1} e^{(\xi+1)^\alpha} \le \alpha e^{(\xi+1)^\alpha} =\alpha\chi(\xi) \le \alpha_0(K)\chi(\xi) \qquad for all\xi\ge 0, \end{align}\] so that since additionally \(\alpha_0(K)\le 1\), we may combine Lemma 23 with (?? ) to see that \[\begin{align} \label{36.2} & &\frac{d}{dt} \int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha} + \gamma(K) \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla v_\varepsilon|^2 + 2 \int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha} \nonumber\\ &\le& \Gamma(K) \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \frac{\chi'^2(w_\varepsilon)+\chi''^2(w_\varepsilon)}{\chi(w_\varepsilon)} \cdot |\nabla w_\varepsilon|^2 + \Gamma(K) \int_{\{\chi''(w_\varepsilon<0\}} |\chi''(w_\varepsilon)| \cdot |\nabla w_\varepsilon|^2 \nonumber\\ & & + \Gamma(K) \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \chi(w_\varepsilon) \qquad for all t>0 and\varepsilon\in (0,1). \end{align}\tag{57}\] Here, (?? ) ensures that \[\begin{align} \label{36.3} \Gamma(K) \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \frac{\chi'^2(w_\varepsilon)+\chi''^2(w_\varepsilon)}{\chi(w_\varepsilon} \cdot |\nabla w_\varepsilon|^2 &\le& 2\Gamma(K) \alpha^2 \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla w_\varepsilon|^2 \nonumber\\ & & + 4e^\frac{2}{\alpha} \Gamma(K) \int_\Omega|\nabla w_\varepsilon|^2 \end{align}\tag{58}\] for all \(t>0\) and \(\varepsilon\in (0,1)\), while from (?? ) we know that \[\label{36.4} \Gamma(K) \int_{\{\chi''(w_\varepsilon<0\}} |\chi''(w_\varepsilon)| \cdot |\nabla w_\varepsilon|^2 \le 4e^\frac{4}{\alpha} \Gamma(K) \int_\Omega|\nabla w_\varepsilon|^2 \qquad for all t>0 and\varepsilon\in (0,1).\tag{59}\] Since (?? ) warrants that \[\begin{align} \Gamma(K) \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot \chi(w_\varepsilon) \le \Gamma(K) \alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + \Big(\frac{e}{\alpha}\Big)^\frac{1}{\alpha} \Gamma(K) |\Omega| \end{align}\] for all \(t>0\) and \(\varepsilon\in (0,1)\), a combination of (57 ) with (58 ) and (59 ) leads to (?? ) with \(\Gamma_2(\alpha,K):= \max\Big\{ 4(e^\frac{2}{\alpha}+ e^\frac{4}{\alpha}) \Gamma(K) \, , \, (\frac{e}{\alpha})^\frac{1}{\alpha} \Gamma(K) |\Omega| \Big\}\). \(\Box\)

6.3 Estimating \(\int_\Omega e^{u^\alpha}\) for small \(\alpha\). Conclusion↩︎

We are thus prepared to establish an approximate version of our main estimate announced in Theorem 1, which indeed can be obtained by combining Lemma 21 with Lemma 25, and estimating the second to last summand in (?? ) by means of the interpolation inequality from Lemma 5.

Proof. We fix \(K>0\) and let \(\alpha_0=\alpha_0(K), \gamma=\gamma(K)\) and \(\Gamma=\Gamma(K)\) from Lemma 23, and abbreviating \[\label{37.02} c_1\equiv c_1(K):=\Lambda_3\big( 2K|\Omega| \big)\tag{60}\] with \(\Lambda_3(\cdot)\) as provided by Lemma 5, we set \[\label{37.2} b\equiv b(K):=\frac{16\Gamma}{D},\tag{61}\] choose \(\alpha=\alpha(K)\in (0,\min\{1,\frac{2}{n}\})\) small enough fulfilling \[\label{37.3} \alpha^2 \le \frac{2D\gamma}{3b(d-D)^2}\tag{62}\] as well as \[\label{37.4} \alpha\le \frac{bD}{8c_1\cdot (3b+\Gamma)},\tag{63}\] and fix \(\varepsilon_0=\varepsilon_0(K)\in (0,1)\) in such a way that \(\varepsilon_0^\alpha\le \frac{\alpha}{2}\). Taking \(\Gamma_2=\Gamma_2(\alpha,K)\) from Lemma 25 and assuming (?? ) as well as (?? ), we may then invoke Lemma 21 along with Lemma 25 to find that \[\begin{align} & &\frac{d}{dt} \bigg\{ b \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + \int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha} \bigg\} + b \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + 2 \int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha} \nonumber\\ & & + \frac{b D \alpha^2}{4} \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla w_\varepsilon|^2 + \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla v_\varepsilon|^2 \nonumber\\ &\le& \frac{3b(d-D)^2 \alpha^2}{2D} \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla v_\varepsilon|^2 + \frac{b(d-D)^2 e^\frac{1}{\alpha}}{2D} \int_\Omega|\nabla v_\varepsilon|^2 \nonumber\\ & & + 3b\alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + 2b\Big(\frac{e}{\alpha}\Big)^\frac{1}{\alpha} \cdot |\Omega| \nonumber\\ & & + 2\Gamma \alpha^2 \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla w_\varepsilon|^2 + \Gamma \alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} \nonumber\\ & & + \Gamma_2 \int_\Omega|\nabla w_\varepsilon|^2 + \Gamma_2 \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0), \end{align}\] where we note that \[\begin{align} \frac{bD\alpha^2}{4} - 2\Gamma\alpha^2 = \frac{bD\alpha^2}{8} \end{align}\] by (61 ), and that \[\begin{align} & &\frac{3b(d-D)^2 \alpha^2}{2D} \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla v_\varepsilon|^2 \nonumber\\ &\le& \frac{3b(d-D)^2 \alpha^2}{2D} \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla v_\varepsilon|^2 \\ &\le& \gamma \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla v_\varepsilon|^2 \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0) \end{align}\] according to (62 ) and the fact that \(\alpha\le 1\). Rearranging and trivially estimating \(2\int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha} \ge \int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha}\) for \(t>0\) and \(\varepsilon\in (0,\varepsilon_0)\), we thus infer that for \[\begin{align} y_\varepsilon(t):=b \int_\Omega\frac{w_\varepsilon+1}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} + \int_\Omega v_\varepsilon^2 e^{(w_\varepsilon+1)^\alpha}, \qquad t\ge 0, \;\varepsilon\in (0,\varepsilon_0), \end{align}\] we have \[\begin{align} \label{37.5} & &y_\varepsilon'(t) + y_\varepsilon(t) + \frac{bD\alpha^2}{8} \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla w_\varepsilon|^2 \nonumber\\ &\le& (3b+\Gamma)\alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} \nonumber\\ & & + c_2 \int_\Omega|\nabla v_\varepsilon|^2 + c_3 \int_\Omega|\nabla w_\varepsilon|^2 + c_4 \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0) \end{align}\tag{64}\] with \(c_2\equiv c_2(K):=\frac{b(d-D)^2 e^\frac{1}{\alpha}}{2D}\), \(c_3\equiv c_3(K):=\Gamma_2\) and \(c_4\equiv c_4(K):=2b(\frac{e}{\alpha})^\frac{1}{\alpha} \cdot |\Omega| + \Gamma_2\).
At this point, based on our restriction that \(\alpha\le \min\{1,\frac{2}{n}\}\) the interpolation result from Lemma 5 applies so as to ensure that, again since \[\begin{align} \int_\Omega w_\varepsilon= \int_\Omega u_\varepsilon+ \int_\Omega v_\varepsilon \le \int_\Omega u_0 + \max \bigg\{ \int_\Omega u_0, \int_\Omega v_0\bigg\} \le 2K|\Omega| \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0) \end{align}\] by (?? ) and (?? ), with \(c_1\) as in (60 ) and with \(c_5\equiv c_5(K):=\Lambda_4(\alpha,2K|\Omega|)\), \(\Lambda_4(\cdot,\cdot)\) being taken from Lemma 5, we have \[\begin{align} & &(3b+\Gamma) \alpha\int_\Omega\frac{(w_\varepsilon+1)^{\alpha+1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} \nonumber\\ &\le& (3b+\Gamma) \alpha\cdot c_1\alpha^2 \int_\Omega\frac{(w_\varepsilon+1)^{2\alpha-1}}{1+\varepsilon w_\varepsilon} \cdot e^{(w_\varepsilon+1)^\alpha} |\nabla w_\varepsilon|^2 \nonumber\\ & & + (3b+\Gamma) \alpha\cdot c_5 \int_\Omega|\nabla w_\varepsilon|^2 + (3b+\Gamma) \alpha\cdot c_5 \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0). \end{align}\] As our smallness condition in (63 ) guarantees that \[\begin{align} (3b+\Gamma) \alpha\cdot c_1\alpha^2 \le \frac{bD\alpha^2}{8}, \end{align}\] this implies that (64 ) entails the inequality \[\begin{align} y_\varepsilon'(t) + y_\varepsilon(t) \le h_\varepsilon(t):=c_2 \int_\Omega|\nabla v_\varepsilon|^2 + \big\{ c_3 + (3b+\Gamma) c_5\alpha\big\} \cdot \int_\Omega|\nabla w_\varepsilon|^2 + c_4 + (3b+\Gamma) c_5\alpha \end{align}\] for all \(t>0\) and \(\varepsilon\in (0,1)\). Since from Lemma 12 we know that with some \(c_6=c_6(K)>0\) we have \[\begin{align} \int_t^{t+1} h_\varepsilon(s) ds \le c_6 \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0), \end{align}\] and since thus \[\begin{align} \int_0^t e^{-(t-s)} h_\varepsilon(s) ds \le \frac{c_6}{1-e^{-1}} \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0) \end{align}\] according to an elementary inequality recorded in [31], this shows that \[\begin{align} y_\varepsilon(t) &\le& y_\varepsilon(0) e^{-t} + \int_0^t e^{-(t-s)} h_\varepsilon(s) ds \\ &\le& b \int_\Omega(u_0+v_0+1) e^{(u_0+v_0+1)^\alpha} + \int_\Omega v_0^2 e^{(u_0+v_0+1)^\alpha} + \frac{c_6}{1-e^{-1}} \\ &\le& b (2K+1) e^{(2K+1)^\alpha} \cdot |\Omega| + K^2 e^{(2K+1)^\alpha} \cdot |\Omega| + \frac{c_6}{1-e^{-1}} \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0). \end{align}\] Observing that \(\frac{\xi+1}{1+\varepsilon\xi} \ge \frac{\varepsilon\xi+1}{1+\varepsilon\xi}=1\) for all \(\xi\ge 0\) and \(\varepsilon\in (0,\varepsilon_0)\) and hence \[\begin{align} y_\varepsilon(t)\ge b\int_\Omega e^{(w_\varepsilon+1)^\alpha} \ge b \int_\Omega e^{u_\varepsilon^\alpha} \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0), \end{align}\] we may conclude as intended. \(\Box\) In its essence, our main result has thereby actually been accomplished already:
Proof  of Theorem 1. Given \(K>0\), from Lemma 26 and Lemma 13 we know that there exist \(c_1=c_1(K)>0\) and \(c_2=c_2(K)>0\) such that whenever (?? ) and (?? ) hold, the solutions \((u_\varepsilon,v_\varepsilon)\) of (23 ) from Lemma 8 satisfy \[\label{14.3} \int_\Omega e^{u_\varepsilon^\alpha(\cdot,t)} \le c_1 \quad and \quad \|v_\varepsilon(\cdot,t)\|_{L^\infty(\Omega)} \le c_2 \qquad for all t>0 and\varepsilon\in (0,\varepsilon_0),\tag{65}\] where \(\alpha=\alpha(K)\) and \(\varepsilon_0=\varepsilon_0(K)\) are as determined by Lemma 26. Apart from that, in view of (?? ), (?? ) and the Fubini-Tonelli theorem, there exists a null set \(N\subset (0,\infty)\) such that with \((u,v)\) and \((\varepsilon_j)_{j\in\mathbb{N}}\) as provided by Lemma 19 we have \[\begin{align} u_\varepsilon(\cdot,t) \to u(\cdot,t) \quad and \quad v_\varepsilon(\cdot,t) \to v(\cdot,t) \quad a.e.~in\Omega \qquad for allt\in (0,\infty)\setminus N \end{align}\] as \(\varepsilon=\varepsilon_j\searrow 0\). By utilizing Fatou’s lemma, from (65 ) we thus infer that \[\begin{align} \int_\Omega e^{u^\alpha(\cdot,t)} \le c_1 \quad and \quad \|v(\cdot,t)\|_{L^\infty(\Omega)} \le c_2 \qquad for allt\in (0,\infty) \setminus N, \end{align}\] so that the claim results upon recalling from Lemma 19 that \((u,v)\) indeed is a global weak solution of (4 ) in the sense specified in Theorem 1. \(\Box\)

7 Appendix: Proof of Lemma 16↩︎

As annonced, let us finally describe how the \(\varepsilon\)-dependent \(W^{1,p}\) bounds claimed in Lemma 16 can be derived from Lemma 14 and Lemma 15.
Proof  of Lemma 16. As an argument addressing a closely related situation can be found detailed in [26], we may confine ourselves here with an outline of the main steps.
Step 1: Deriving an enery-type inequality for \(\int_\Omega|\nabla v_\varepsilon|^p\) with \(p\ge 4\). Using the identities \[\label{7.11} \nabla u_\varepsilon=\nabla w_\varepsilon-\nabla v_\varepsilon\quadand\quad \nabla\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}=\frac{1}{(1+\varepsilon u_\varepsilon)^2} \nabla u_\varepsilon\tag{66}\] as well as the inequality \[\label{7.12} \Big|\frac{u_\varepsilon}{1+\varepsilon u_\varepsilon}\Big| \le \frac{1}{\varepsilon}\tag{67}\] as seen in [26] we can obtain that for each \(p\ge 4\) and any \(\sigma>0\) one can find \(K_1(\sigma, \varepsilon, p, u_0, v_0)>0\) satisfying \[\label{7.13} \frac{1}{p}\frac{d}{dt} \int_\Omega|\nabla v_\varepsilon|^p +\frac{d}{16} \int_\Omega|\nabla v_\varepsilon|^{p-2}|D^2 v_\varepsilon|^2 \le \sigma \int_\Omega|\nabla w_\varepsilon|^{p+2} +K_1\int_\Omega|\nabla v_\varepsilon|^{p+2}+K_1\tag{68}\] for all \(t\in (0,T_{max,\varepsilon})\).
Step 2: Establishing an enery-type inequality for \(\int_\Omega|\nabla w_\varepsilon|^p\) with \(p\ge 4\). Relying on (?? ) and (66 ) and following [26], we can show that for all \(p\ge 4\) there exists \(K_2(\varepsilon, p, u_0, v_0)>0\) satisfying \[\label{7.14}\frac{1}{p}\frac{d}{dt} \int_\Omega|\nabla w_\varepsilon|^p +\frac{D}{2} \int_\Omega|\nabla w_\varepsilon|^{p-2}|D^2 w_\varepsilon|^2 \le K_2 \int_\Omega|\nabla w_\varepsilon|^{p-2}|D^2 v_\varepsilon|^2 +K_2\int_\Omega|\nabla w_\varepsilon|^{p} +K_2\int_\Omega|\nabla v_\varepsilon|^{p}\tag{69}\] for all \(t\in (0,T_{max,\varepsilon})\).
Step 3: Studying the evolution of the coupled-gradient functional \(\int_\Omega|\nabla v_\varepsilon|^2 |\nabla w_\varepsilon|^{p-2}\) for \(p\ge 6\). To control the first integral on the right-hand side of (69 ), using (66 ) and (67 ) we see that for each \(p\ge 6\) and any \(\eta>0\) there exists \(K_3(\eta, \varepsilon, p, u_0, v_0)>0\) fulfilling \[\begin{align} \label{7.15} \frac{d}{dt} \int_\Omega|\nabla v_\varepsilon|^2 |\nabla w_\varepsilon|^{p-2} &+& \frac{d}{4}\int_\Omega|\nabla w_\varepsilon|^{p-2}|D^2 v_\varepsilon|^2 \nonumber\\ &\le& \eta \int_\Omega|\nabla w_\varepsilon|^2 |D^2w_\varepsilon|^2 + \eta\int_\Omega|\nabla w_\varepsilon|^{p+2}\nonumber\\ & & +K_3 \int_\Omega|\nabla v_\varepsilon|^{p-2}|D^2v_\varepsilon|^2 +K_3\int_\Omega|\nabla v_\varepsilon|^{p+2} +K_3 \end{align}\tag{70}\] for all \(t\in (0,T_{max,\varepsilon})\) (cf. [26]).
Step 4: Recalling two useful interpolation inequalities. In order to expediently deal with the integrals \(\int_\Omega|\nabla w_\varepsilon|^{p+2}\) and \(\int_\Omega|\nabla v_\varepsilon|^{p+2}\) appearing on the right-hand sides of (68 )-(70 ), we shall invoke the following two interpolation properties (cf. [26]):
i) Given any \(p\ge 2\), one can find \(K_{41}>0\) such that whenever \(\varphi\in C^2(\overline{\Omega})\) satisfies \(\frac{\partial\varphi}{\partial\nu}=0\) on \(\partial\Omega\), we have \[\label{7.16} \int_\Omega|\nabla\varphi|^{p+2} \le K_{41} \cdot \bigg\{ \int_\Omega|\nabla\varphi|^{p-2} |D^2 \varphi|^2 \bigg\} \cdot \|\varphi\|_{L^\infty(\Omega)}^2.\tag{71}\] ii) Let \(\omega:(0,\infty)\to (0,\infty)\) be nondecreasing, and let \(p\ge 2\) and \(\tilde{\eta}>0\). Then there exists \(K_{42}(\tilde{\eta},q,\omega)>0\) such that if \(\varphi\in C^2(\overline{\Omega})\) is such that \(\frac{\partial\varphi}{\partial\nu}=0\) on \(\partial\Omega\) and that \[\begin{align} \qquad for each \delta>0 and any x\in\overline{\Omega} and y\in\overline{\Omega} fulfilling |x-y|<\omega(\delta), we have|\varphi(x)-\varphi(y)| < \delta, \end{align}\] it follows that \[\label{7.17} \int_\Omega|\nabla\varphi|^{p+2} \le \tilde{\eta} \int_\Omega|\nabla\varphi|^{p-2} |D^2\varphi|^2 + K_{42}(\tilde{\eta},p,\omega) \|\varphi\|_{L^\infty(\Omega)}^{p+2}.\tag{72}\] Step 5: Completing the proof. To compensate the summands appearing on the right-hand sides of (68 )-(70 ) by means of the diffusion-related integrals on the left-hand sides therein, we need to suitably select the two free small paramters \(\eta\) and \(\sigma\) in (70 ) and (68 ) and design an appropriate linear combination of the functionals \(\int_\Omega|\nabla v_\varepsilon|^p, \int_\Omega|\nabla w_\varepsilon|^p\) and \(\int_\Omega|\nabla v_\varepsilon|^2 |\nabla w_\varepsilon|^{p-2}\). For this purpose, we first invoke (71 ) in conjunction with Lemma 15 and Lemma 13 to fix \(c_1\equiv c_1(\varepsilon, p, u_0, v_0)\) such that \[\label{7.18} \int_\Omega|\nabla w_\varepsilon|^{p+2} \le c_1 \int_\Omega|\nabla w_\varepsilon|^{p-2} |D^2w_\varepsilon|^2 \qquadfor all t\in (0,T_{max,\varepsilon}) and \varepsilon\in (0,1),\tag{73}\] and taking \(K_2 =K_2(\varepsilon, p, u_0, v_0)\) as obtained in (69 ), we let \[\label{7.19} \beta=\beta(\varepsilon, p, u_0, v_0) :=\frac{d}{4K_2}\tag{74}\] as well as \[\label{7.110} \eta =\eta(\varepsilon, p, u_0, v_0) :=\min\Big\{\frac{\beta D}{4}, \, \, \frac{\beta D}{8c_1}\Big\}.\tag{75}\] Thereupon fixing \(K_3=K_3(\eta, \varepsilon, p, u_0, v_0)\) such that (70 ) holds, we take \[\label{7.111} b\equiv b(\varepsilon, p, u_0, v_0) :=\frac{d}{32 K_3}\tag{76}\] and \[\label{7.112} \sigma\equiv \sigma(\varepsilon, p, u_0, v_0) :=\frac{\beta b D}{16 c_1}\tag{77}\] and let \(K_1=K_1(\sigma, \varepsilon, p, u_0, v_0)\) be as accordingly be introduced near (68 ). Now defining \[\label{7.113} y_\varepsilon(t) :=\frac{1}{p} \int_\Omega|\nabla v_\varepsilon(\cdot, t)|^p +b\int_\Omega|\nabla v_\varepsilon(\cdot, t)|^2 |\nabla w_\varepsilon(\cdot, t)|^{p-2} +\frac{\beta b}{p}\int_\Omega|\nabla w_\varepsilon(\cdot, t)|^p, \qquad t\in [0,T_{max,\varepsilon}),\tag{78}\] by straightforward computation of (68 )-(70 ) with (74 )-(78 ) we obtain that \[\begin{align} \label{7.114}y_\varepsilon'(t) &=& -\Big(\frac{d}{16}-bK_3\Big) \int_\Omega|\nabla v_\varepsilon|^{p-2} |D^2v_\varepsilon|^2 -\Big(\frac{bd}{4}-\beta b K_2\Big) \int_\Omega|\nabla w_\varepsilon|^{p-2} |D^2 v_\varepsilon|^2\nonumber\\ & & -\Big(\frac{\beta bD}{2}-b\eta\Big)\int_\Omega|\nabla w_\varepsilon|^{p-2} |D^2w_\varepsilon|^2\nonumber\\ & & +(K_1+bK_3) \int_\Omega|\nabla v_\varepsilon|^{p+2} +(\sigma +b\eta)\int_\Omega|\nabla w_\varepsilon|^{p+2}\nonumber\\ & & + \beta b K_2 \int_\Omega|\nabla v_\varepsilon|^p + \beta b K_2 \int_\Omega|\nabla w_\varepsilon|^p \nonumber\\ & & +K_1 +bK_3\nonumber\\ &\le& -\frac{d}{32} \int_\Omega|\nabla v_\varepsilon|^{p-2} |D^2v_\varepsilon|^2 -\frac{\beta b D}{4} \int_\Omega|\nabla w_\varepsilon|^{p-2} |D^2w_\varepsilon|^2 \nonumber\\[1mm] & & +I \qquadfor all t\in (0,T_{max,\varepsilon}) \end{align}\tag{79}\] due to the fact that \(\frac{d}{16}-bK_3=\frac{d}{32}, \frac{bd}{4}-\beta b K_2=0\) and \(\frac{\beta bD}{2}-b\eta\ge \frac{\beta bD}{4}\) by (76 ), (74 ) and the first restriction in (75 ), respectively, where for \(t\in (0,T_{max,\varepsilon})\) we have set \[\begin{align}I &:=& (K_1+bK_3) \int_\Omega|\nabla v_\varepsilon|^{p+2} +(\sigma +b\eta)\int_\Omega|\nabla w_\varepsilon|^{p+2}\nonumber\\ & & + \beta b K_2 \int_\Omega|\nabla v_\varepsilon|^p + \beta b K_2 \int_\Omega|\nabla w_\varepsilon|^p +K_1 +bK_3. \end{align}\] Here, Young’s inequality entails that for all \(t\in (0,T_{max,\varepsilon})\), \[\label{7.116} \beta b K_2 \int_\Omega|\nabla v_\varepsilon|^p \le \beta b K_2 \int_\Omega|\nabla v_\varepsilon|^{p+2} + \beta b K_2 |\Omega|\tag{80}\] and \[\begin{align} \label{7.117} \beta b K_2 \int_\Omega|\nabla w_\varepsilon|^p &=& \int_\Omega\Big(\sigma |\nabla w_\varepsilon|^{p+2}\Big)^\frac{p}{p+2} \cdot \sigma^{-\frac{p}{p+2}} \beta b K_2 \nonumber\\ &\le& \sigma \int_\Omega|\nabla w_\varepsilon|^{p+2} +c_2 \end{align}\tag{81}\] with \(c_2\equiv c_2(\varepsilon, p, u_0, v_0):=\sigma^{-\frac{p}{2}}\big(\beta b K_2\big)^\frac{p+2}{2}\), and from (73 ) we obtain that \[\label{7.118} (2\sigma +b\eta)\int_\Omega|\nabla w_\varepsilon|^{p+2} \le (2\sigma +b\eta) c_1\int_\Omega|\nabla w_\varepsilon|^{p-2}|D^2w_\varepsilon|^2 \qquad for allt\in (0,T_{max,\varepsilon}).\tag{82}\] Collecting (80 )-(82 ) leads to the inequality \[\label{7.119} I \le \Big(K_1+bK_3+\beta bK_2 +\frac{1}{p}+b\Big) \int_\Omega|\nabla v_\varepsilon|^{p+2} + (2\sigma +b\eta) c_1\int_\Omega|\nabla w_\varepsilon|^{p-2}|D^2w_\varepsilon|^2 +c_3 \qquad for allt\in (0,T_{max,\varepsilon})\tag{83}\] with \(c_3\equiv c_3(\varepsilon, p, u_0, v_0):=K_1+bK_3 + \beta b K_2 |\Omega| + c_2\), and inserting this into (79 ) we arrive at the inequality \[\begin{align} \label{7.120} y_\varepsilon'(t) + \frac{d}{32} \int_\Omega|\nabla v_\varepsilon|^{p-2} |D^2v_\varepsilon|^2 &\le& -\Big\{\frac{\beta bD}{4} -(2\sigma +b\eta)c_1\Big\} \int_\Omega|\nabla w_\varepsilon|^{p-2} |D^2w_\varepsilon|^2\nonumber\\ & & + (K_1+bK_3+\beta bK_2) \int_\Omega|\nabla v_\varepsilon|^{p+2} +c_3\nonumber\\ &\le & (K_1+bK_3+\beta bK_2) \int_\Omega|\nabla v_\varepsilon|^{p+2} +c_3 \end{align}\tag{84}\] for all \(t\in (0,T_{max,\varepsilon})\), because \(\frac{\beta bD}{4} -(2\sigma +b\eta)c_1 =\big(\frac{\beta bD}{8} -2\sigma c_1\big) + \big(\frac{\beta bD}{8} -b\eta c_1\big) =\frac{\beta bD}{8} -b\eta c_1 \ge 0\) according to (77 ) and the second restriction in (75 ). Now in line with Lemma 14, we may apply (72 ) to \(\tilde{\eta}:=\frac{1}{K_1+bK_3+\beta bK_2}\cdot \frac{d}{32}\) to find \(c_4=c_4(\varepsilon, p, u_0, v_0)>0\) such that \[\begin{align} (K_1+bK_3+\beta bK_2) \int_\Omega|\nabla v_\varepsilon|^{p+2} \le \frac{d}{32} \int_\Omega|\nabla v_\varepsilon|^{p-2} |D^2v_\varepsilon|^2 +c_4, \end{align}\] so that (84 ) implies that \[\begin{align} y_\varepsilon'(t) \le c_3+c_4 \qquadfor all t\in (0,T_{max,\varepsilon}) \end{align}\] and thus \(y_\varepsilon(t) \le y_\varepsilon'(0) \cdot (c_3+c_4) T_{max,\varepsilon}\) for all \(t\in (0,T_{max,\varepsilon})\). Since \(|\nabla u_\varepsilon(\cdot, t)|^p \le 2^{p-1} (|\nabla w_\varepsilon(\cdot, t)|^p + |\nabla v_\varepsilon(\cdot, t)|^p)\), this leads to (?? ). \(\Box\)

Acknowledgement. The first author was supported by the National Natural Science Foundation of China (No. 12571222). The second author acknowledges support of the Deutsche Forschungsgemeinschaft (Project No. 462888149).

References↩︎

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1. taoys@sjtu.edu.cn↩︎

2. michael.winkler@math.uni-paderborn.de↩︎